Function and Operator Theory on Large Bergman Spaces

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Function and Operator Theory on Large Bergman spaces

Hicham Arroussi

A questa tesi doctoral està subjecta a la llicència Reconeixement 3.0. Es panya de Creative Commons .

Esta tesis doctoral está sujeta a la licencia Reconocimi ento 3.0. España de Creative Commons .

Th is doctoral thesis is licensed under the Creative Commons Att ribution 3.0. Spain License .

Function and Operator Theory on Large Bergman spaces

Hicham Arroussi

Programa de Doctorat en Matem`atiquesi Inform`atica Universitat of Barcelona

A thesis submitted for the degree of Doctor in Mathematics (PhD)

Advisor: Dr. Jordi Pau Plana

Declaration

Hicham Arroussi wrote this thesis for the award of Doctor of Mathematics under the supervision of Jordi Pau Plana, and Joaquim Ortega-Cerd`aas a tutor.

Dr. Jordi Pau Plana Barcelona, February 2016

iii iv Acknowledgements

First, I would like to express my sincerest thanks to my supervisor Dr. Jordi Pau. Through- out this project his enthusiasm for the subject matter was a great source of inspiration. In addition to this, his excellent supervision kept me on track yet still allowed me to produce my best work.

I must also thank the staff of the Department of Applied Mathematics and Analysis who have made this time so enjoyable and for being there whenever I needed help, especially Joaquim Ortega-Cerdà,Xavier Massaneda, Daniel Pascuas, Joan Fàbrega,Fina Casasayas, Maria J. Carro, Javier Soria, Oriol Pujol, and Konstantin Dyakonov with whom I have shared the office. I am also extremely grateful to Josep Vives from the Department of Probability, Logic and Statistics. All of them were available when I wanted to talk maths as well as those times when I would have preferred to speak about anything else.

To my close friends and family I will be eternally grateful. Without your encourage- ment I would never have completed this work. You will never know how much your love, kindness and support has meant to me these past years and I just hope that I can pay it all back in the years to come.

v vi Contents

Introduction 1

1 Large Bergman spaces 5 1.1 Basic properties ...... 5 1.2 The class of weights W and examples ...... 12 1.3 Test functions and some estimates ...... 13 1.4 Carleson type measures ...... 21 1.5 Atomic Decomposition ...... 23

2 Toeplitz operators 29 2.1 Boundedness and compactness ...... 30 2.2 Membership in Schatten classes ...... 32

3 Area operators 45 3.1 The case p ≤ q ...... 46 3.2 The case q < p ...... 49

4 The exponential weight 53 4.1 Integral estimates for reproducing kernels ...... 53 4.2 Bounded projections ...... 54 4.3 Complex interpolation and duality ...... 57 4.4 Atomic decomposition ...... 62 4.5 Toeplitz operators ...... 64 4.6 Hankel operators ...... 73 4.7 Examples of weights in the class E ...... 80

5 Open questions and future research 97 5.1 Extension of the results to p 6=2...... 97 5.2 Localization and Compactness ...... 98 5.3 Big Hankel operators ...... 100 5.4 Small Hankel operators ...... 101

Bibliography 103

vii viii Introduction

The theory of Bergman spaces has been a central subject of study in complex analysis during the past decades. The book [7] by S. Bergman contains the first systematic treat- ment of the Hilbert space of square integrable analytic functions with respect to Lebesgue area measure on a domain. His approach was based on a reproducing kernel that became known as the Bergman kernel function. When attention was later directed to the spaces Ap over the unit disk, it was natural to call them Bergman spaces. As counterparts of Hardy spaces, they presented analogous problems. However, although many problems in Hardy spaces were well understood by the 1970s, their counterparts for Bergman spaces were generally viewed as intractable, and only some isolated progress was done. The 1980s saw the emerging of operator theoretic studies related to Bergman spaces with important contributions by several authors. Their achievements on Bergman spaces with standard weights are presented in Zhu’s book [77]. The main breakthroughs came in the 1990s, where in a flurry of important advances, problems previously considered intractable began to be solved. First came Hedenmalm‘s construction of canonical divisors [26], then Seip’s description [59] of sampling and interpolating sequences on Bergman spaces, and later on, the study of Aleman, Richter and Sundberg [1] on the invariant subspaces of A2, among others. This attracted other workers to the field and inspired a period of intense research on Bergman spaces and related topics. Nowadays there are rich theories on Bergman spaces that can be found on the textbooks [27] and [22].

Meanwhile, also in the nineties, some isolated problems on Bergman spaces with exponential type weights began to be studied. These spaces are large in the sense that they contain all the Bergman spaces with standard weights, and their study presented new dif- ﬁculties, as the techniques and ideas that led to success when working on the analogous problems for standard Bergman spaces, failed to work on that context. It is the main goal of this work to do a deep study of the function theoretic properties of such spaces, as well as of some operators acting on them. It turns out that large Bergman spaces are close in spirit to Fock spaces [79], and many times mixing classical techniques from both Bergman and Fock spaces in an appropriate way, can led to some success when studying large Bergman spaces.

This dissertation is structured into ﬁve chapters.

1 In Chapter 1, we give some basic properties of the large Bergman spaces Ap(ω), for weights ω in a certain class W of rapidly decreasing weights, considered previously in [23] and [51]. The concrete deﬁnition of the class is given, and it is seen that contains the family of exponential type weights −A ω (z) = exp , σ > 0, A > 0, (0.1) σ (1 − |z|2)σ and also weights decreasing even faster such as some double exponential weights. We collect some of the well known results proved in earlier works (pointwise estimates, completeness, construction of suitable test functions, description of Carleson-type measures, etc.), but we also include some new ones as the atomic decomposition for the Hilbert space A2(ω), or the estimates for the solutions of the d-bar equation obtained in Theorem 1.4, a result used later in Chapter 4 in order to characterize bounded and compact Hankel operators with conjugate analytic symbols. The test functions given in Lemma C are useful in order to study several problems on Ap(ω) even for p 6= 2 (see [9] for the study of sampling and interpolating sequences, or [51] for the study of the boundedness of certain Cesaro type integration operators acting between large weighted Bergman spaces), but the fact that estimates for the norm of the reproducing kernels Kz are only available when p = 2 allows us to consider other problems only in the Hilbert space case, as for example, the study of Toeplitz operators in the next chapter. The topics of this chapter will serve as a basis for our later chapters.

Chapter 2 is devoted to characterize the boundedness, compactness and membership 2 in Schatten classes of the Toeplitz operator Tµ acting on A (ω) for weights in our class W. If µ is a finite positive Borel measue on D, the Toeplitz operator Tµ is the integral type operator given by Z Tµf(z) = f(ξ)Kz(ξ) ω(ξ) dµ(ξ), f ∈ H(D), D where Kz denotes the reproducing kernel at the point z ∈ D. The operator Tµ acting on standard weighted Bergman spaces has been extensively studied [75]. Luecking was probably one of the first who considered Tµ with measures as symbols, and the study of Toeplitz operators acting on large weighted Bergman spaces was initiated by Lin and Rochberg [35], where they proved descriptions of the boundedness and compactness of the Toeplitz operator in terms of the behavior of a certain averaging function of µ. Since their class of weights is slightly different of our class, we will offer a proof for the both descriptions in Theorem 2.1. The main result of this chapter is Theorem 2.8 where a complete description, valid for all 2 0 < p < ∞, of when the Toeplitz operator Tµ acting on A (ω) belongs to the Schatten ideal Sp. This extends the description obtained for standard Bergman spaces [78], and solves a problem posed by Lin and Rochberg in [32], where only one implication was proved. Just to mention here that the results of this chapter has been recently published in our paper [4].

2 In Chapter 3 we study the area operator Aµ acting on large weighted Bergman spaces. Given a positive Borel measure µ on the unit disk D, the area operator Aµ is the sublinear operator deﬁned by Z dµ(z) Aµ(f)(ζ) = |f(z)| 2 , ζ ∈ T := ∂D, Γ(ζ) 1 − |z|

where Γ(ζ) is a typical non-tangential approach region (Stolz region) with vertex ζ ∈ T. The area operator is useful in harmonic analysis, and is closely related to, for example, non-tangential maximal functions, Poisson integrals, Littlewood-Paley operators, tent spaces, etc. The study of the area operator acting on the classical Hardy spaces Hp was initiated by W. Cohn [14]. He proved that, for 0 < p < ∞, the area operator p p Aµ : H → L (T) is bounded if and only if µ is a classical Carleson measure. This was p q pursued later in [24], where a full description of the boundedness of Aµ : H → L (T) for the case 0 < p < q < ∞ and 1 ≤ q < p < ∞ was obtained. In the setting of standard p Bergman spaces Aα, the study of the area operator was initiated in [69] by Z. Wu, who p q obtained a characterization of the boundedness of Aµ : Aα → L (T) for 1 ≤ p, q < ∞. In this chapter we are going to extend these results to our large Bergman spaces Ap(ω) for weights ω in the class W and characterize those positive Borel measures µ for which the p q area operator Aµ : A (ω) → L (T), 1 ≤ p, q < ∞ is bounded. Some of the key tools used in the proofs are: the test functions given in Lemma C, the classical description of Carleson measures for Hardy spaces, and the recent description [51] of Carleson type measures for large weighted Bergman spaces.

In Chapter 4 we add an extra condition to our class of weights W, and consider the class E that consists of those weights ω ∈ W satisfying Z 1/2 −1/2 |Kz(ξ)| ω(ξ) dA(ξ) ≤ C ω(z) , z ∈ D. (0.2) D This condition will allow us to extend some of the previous results to the non-Hilbert space setting. It has been recently proved in [16] that the exponential type weights ωσ given in (0.1) with σ = 1 satisfy the previous condition and, therefore, they are in the class E. We are able to show, following the proof given in [16] with non-trivial modiﬁcations, that all the family of exponential type weights given in (0.1) are in the class E for all 0 < σ < ∞. The integral estimate (0.2) allows to study other properties and operators, such as the Bergman projection which is given by Z Pω(f)(z) = f(ξ) Kz(ξ) ω(ξ) dA(ξ), z ∈ D. D 2 The boundedness of the Bergman projection Pω on L (ω) is trivial from the general theory of Hilbert spaces. In contrast with the case of standard Bergman spaces (where the Bergman projection is bounded for 1 < p < ∞), in the case of exponential type weights it turns out that the the natural Bergman projection is not bounded on Lp(ω) unless p = 2

3 (see [18] and [71]). At first glance, this may look as a surprise, but when one takes into account the similarities with Fock spaces, this seems to be more natural. It turns out that, similarly as in the setting of Fock spaces, when studying problems where the reproducing kernels are involved, the most convenient setting are the spaces Ap(ωp/2). As a consequence of condition (0.2), we get the right estimates for the norm of the reproducing kernels in Ap(ωp/2) for 1 ≤ p ≤ ∞. Also, we prove in Theorem 4.4 that, for weights in the class E, p p/2 p p/2 the Bergman projection Pω : L (ω ) → A (ω ) is bounded for 1 ≤ p ≤ ∞. A consequence of that result will be the identification of the dual space of Ap(ωp/2) with the space p0 p0/2 1 1/2 ∞ 1/2 0 A (ω ) and A (ω ) with A (ω ) under the natural integral pairing h·, ·iω, where p denotes the conjugate exponent of p. Afterwards, by using the duality and the estimates for the p-norms of reproducing kernels, we extend in Theorem 4.12 the atomic decomposition obtained in Chapter 1 to the non-Hilbert space setting: for weights ω ∈ E, every function in the weighted Bergman space Ap(ωp/2), 1 ≤ p < ∞ can be decomposed into a series of very nice functions (called atoms). These atoms are defined in terms of kernels functions and in some sense act as a basis for the space Ap(ωp/2). The atomic decomposition for Bergman space with standard weights was obtained by Coifman and Rochberg [15], and has become a powerful tool in the study of the properties of weighted Bergman spaces having found many applications. The norm estimates for the reproducing kernels in Ap(ωp/2) for 1 ≤ p < ∞ permits to extend the results on the boundedness and compactness of Toeplitz operators Tµ and to consider the action of Tµ between different large weighted Bergman spaces, and to find p p/2 q q/2 a general description of when Tµ : A (ω ) → A (ω ) is bounded or compact for all values of 1 ≤ p, q < ∞. Furthermore, we also generalize the results obtained in [23] on the boundedness and compactness of big Hankel operators Hg with conjugate analytic symbols to the non-Hilbert space setting, characterizing for all 1 < p, q < ∞, when p p/2 q q/2 Hg : A (ω ) → L (ω ), is bounded or compact. As mentioned earlier, one of the key tools used is the estimates for the d-bar equation obtained in the first chapter.

Finally, in Chapter 5, we discuss some problems we have not been able to solve, as well other interesting problems to look on the future.

4 Chapter 1

Large Bergman spaces

1.1 Basic properties

dxdy Let D be the unit disk in the complex plane, dA(z) = π be the normalized area measure on D, and let H(D) denote the space of all analytic functions on D. A weight is a positive function ω ∈ L1(D, dA). When ω(z) = ω(|z|) for all z ∈ D, we say that ω is radial. For 0 < p < ∞, the weighted Bergman space Ap(ω) is the space of all functions f ∈ H(D) such that 1 Z p p kfkAp(ω) = |f(z)| ω(z) dA(z) < ∞. D Our main goal is to study the Bergman spaces Ap(ω) for a large class of weights, which includes certain rapidly radial decreasing weights, that is, weights that are going to decrease faster than any standard weight (1 − |z|2)α, α > 0, such as the exponential type weights −A ω (z) = exp , σ > 0, A > 0. (1.1) σ (1 − |z|2)σ Deﬁnition 1.1. We say that a positive function τ belongs to the class L if it satisﬁes the following two properties: (A) there exists c1 > 0 such that τ(z) ≤ c1(1 − |z|), for all z ∈ D. (B) there exists c2 > 0 such that |τ(z) − τ(w)| ≤ c2|z − w|, for all z, w ∈ D.

For a ∈ D and δ > 0 we use too the notation D(δτ(a)) for the euclidian disc centered at a and radius δτ(a) and min(1, c ,−1 c−1) m := 1 2 , τ 4 where c1 and c2 are the constants appearing in the previous definition. From the above definition it is easy to show (see [51, Lemma 2.1]) that if τ ∈ L and z ∈ D(δτ(a)), then 1 τ(a) ≤ τ(z) ≤ 2 τ(a), (1.2) 2 5 for sufficiently small δ > 0, that is, for δ ∈ (0, mτ ). This fact will be used repeatedly throughout this work.

Also, throughout this manuscript, the letter C will denote an absolute constant whose value may change at diﬀerent occurrences. We also use the notation a . b to indicate that there is a constant C > 0 with a ≤ Cb, and the notation a b means that a . b and b . a.

Deﬁnition 1.2. We say that a weight ω is in the class L∗ if it is of the form ω = e−2ϕ, where ϕ ∈ C2(D) with ∆ϕ > 0, and (∆ϕ(z))−1/2 τ(z), with τ(z) being a function in the class L. Here ∆ denotes the classical Laplace operator.

The following result from [51, Lemma 2.2] will play an essential role in the proof of the main theorems of this work and can be thought as some type of generalized sub-mean value property for |f|p ω that gives the boundedness of the point evaluation functionals on Ap(ω).

Lemma A. Let ω ∈ L∗, 0 0 suﬃciently small. It can be seen from the proof given in [51] that one only needs f to be holomorphic in a neighbourhood of D(δτ(z)).

Another consequence of the above result is that the Bergman space Ap(ωβ) is a Banach space when 1 ≤ p < ∞ and a complete metric space when 0 < p < 1. In particular A2(ω) is a Hilbert space with the inner product inherited from L2(D, ω dA). Another immediate consequence is that, for any z ∈ D, the point evaluations `z : f 7−→ f(z) are bounded linear functionals on Ap(ωβ). In particular, when p = 2, it follows from the Riesz repre- 2 sentation theorem that there are functions Kz ∈ A (ω) with k`zk = kKzkA2(ω) such that Z f(z) = hf, Kziω := f(ξ)Kz(ξ) ω(ξ)dA(ξ) D 2 for all f ∈ A (ω). The function Kz has the property that Kz(ξ) = Kξ(z), and is called the reproducing kernel for the Bergman space A2(ω). It is straightforward to see from the previous formula that the orthogonal (Bergman) projection from L2(D, ωdA) to A2(ω) is given by Z Pωf(z) = f(ξ)Kz(ξ) ω(ξ) dA(ξ). D We also need a similar estimate as in Lemma A, but for the gradient of |f| ω1/2.

6 ∗ Lemma 1.1. Let ω ∈ L and 0 0 suﬃciently small there exists a constant C(δ0) > 0 such that

C(δ ) Z 1/p ∇(|f| ω1/2)(z) ≤ 0 |f(ξ)|p ω(ξ)p/2dA(ξ) , 1+ 2 τ(z) p D(δ0τ(z)/2) for all f ∈ H(D). Proof. We follow the method used in [49]. Without loss of generality we can assume z = 0. Then, applying the Riesz’s decomposition (see for example [56]) of the subharmonic δ0 function ϕ in D(0, 2 τ(0)), we obtain Z ϕ(ξ) = u(ξ) + G(ξ, η)∆ϕ(η)dA(η), (1.3) r D( 2 )

r where r = δ0τ(0), u is the least harmonic majorant of ϕ in D(0, 2 ) and G is the Green function deﬁned for every ξ, η ∈ D(0, r), ξ 6= η by

2 r(ξ − η) G(ξ, η) := log . r2 − ηξ¯

r For ξ, η ∈ D(0, 2 ) we have

2 2 ∂G r − |η| (ξ, η) = ∂ξ |ξ − η||r2 − ηξ¯| r2 4 ≤ ≤ . |ξ − η| · r2 − |η||ξ| 3|ξ − η| Then, Z ∂ϕ(0) ∂u(0) ∂G − ≤ (0, η) ∆ϕ(η)dA(η) ∂ξ ∂ξ r ∂ξ D( 2 ) (1.4) Z 1 dA(η) δ0 . 2 = . τ(0) r |η| τ(0) D( 2 )

We pick a function h ∈ H(D) such that Re(h) = u. Also,

0 −ϕ 1 f (ξ)f(ξ) ∂ϕ −ϕ(ξ) ∇(|f|e )(ξ) = − (ξ)|f(ξ)| e 2 |f(ξ)| ∂ξ

2 1 0 |f(ξ)| ∂ϕ −ϕ(ξ) = f (ξ) − 2 (ξ) e . 2 f(ξ) ∂ξ

7 0 ∂u Therefore, since h (0) = 2 ∂ξ (0), we get

−ϕ 1 0 ∂ϕ −ϕ(0) ∇(|f|e )(0) = f (0) − 2f(0) (0) e 2 ∂ξ

1 0 ∂u −ϕ(0) ∂u ∂ϕ −ϕ(0) ≤ f (0) − 2f(0) (0) e + (0) − (0) |f(0)|e 2 ∂ξ ∂ξ ∂ξ

−h ∂(fe )(0) u(0)−ϕ(0) ∂u ∂ϕ −ϕ(0) e + (0) − (0) |f(0)|e . . ∂ξ ∂ξ ∂ξ

By (1.4) we have

∂u ∂ϕ −ϕ(0) δ0 −ϕ(0) (0) − (0) |f(0)|e |f(0)| e . ∂ξ ∂ξ . τ(0) This gives −h −ϕ ∂(fe )(0) u(0)−ϕ(0) |f(0)| −ϕ(0) ∇(|f|e )(0) e + e . (1.5) . ∂ξ τ(0) It follows from Lemma A that !1/p |f(0)| 1 Z e−ϕ(0) |f(z)|pe−pϕ(z)dA(z) . (1.6) . 1+ 2 τ(0) τ(0) p D(δ0τ(0)/2)

To manage the other term appearing in (1.5), notice that if we use the identity (1.3) with the function φ(ξ) = |ξ|2 − (r/2)2 (since ∆φ(ξ) = 4 and its least harmonic majorant is uφ = 0), we obtain Z 1 G(ξ, η) dA(η) = |ξ|2 − (r/2)2. r 4 D( 2 ) 1 1 Therefore, since ∆ϕ(η) = τ(η)2 . τ(0)2 = ∆ϕ(0) and the Green’s function G ≤ 0, we obtain r for every ξ ∈ D(0, 2 ) Z u(ξ) − ϕ(ξ) = − G(ξ, η)∆ϕ(η)dA(η) r D( 2 )

∆ϕ(0) 1 (r/2)2 − |ξ|2 = (r/2)2 − |ξ|2. . 4 4τ(0)2

This gives u(0)−ϕ(0) Cδ2 e ≤ e 0 .

8 Therefore −h −h ∂(fe )(0) u(0)−ϕ(0) ∂(fe )(0) e . (1.7) ∂ξ . ∂ξ On the other hand, using Cauchy’s inequality, the fact that ϕ − u ≤ 0 and Lemma A, we get

−h Z −h(η) ∂(fe ) f(η)e (0) . 2 dη ∂ξ δ0τ(0) η |η|= 4

Z 1 −ϕ(η) ϕ(η)−u(η) . 2 2 |f(η)|e e |dη| δ τ(0) δ0τ(0) 0 |η|= 4

Z Z !1/p 1 1 p −pϕ(z) . 2 2 |f(z)| e dA(z) |dη|. τ(0) δ0τ(0) τ(η) |η|= 4 D(δ0τ(η)/4) Finally, using τ(η) τ(0), we obtain

1/p −h Z Z ! ∂fe 1 1 p −pu(z) (0) . 2 2 |f(z)| e dA(z) |dη| ∂ξ τ(0) δ0τ(0) τ(0) |η|= 4 D(δ0τ(0)/2)

!1/p 1 Z |f(z)|pe−pϕ(z)dA(z) . . 1+ 2 τ(0) p D(δ0τ(0)/2)

Bearing in mind (1.7) this gives

!1/p ∂(fe−h)(0) 1 Z eu(0)−ϕ(0) |f(z)|pe−pϕ(z)dA(z) . . 1+ 2 ∂ξ τ(0) p D(δ0τ(0)/2)

Putting this and (1.6) into (1.5) we get the result. We shall also need the following lemma on coverings due to Oleinik [47]

Lemma B. Let τ ∈ L and δ ∈ (0, mτ ). Then there exists a sequence of points {zj} ⊂ D, such that the following conditions are satisﬁed:

(1) zj ∈/ D(δτ(zk)), j 6= k. S (2) j D(δτ(zj)) = D S (3) De(δτ(zj)) ⊂ D(3δτ(zj)), where De(δτ(zj)) = D(δτ(z)) z∈D(δτ(zj )) j = 1, 2, 3, ...

9 (4) {D(3δτ(zj))} is a covering of D of ﬁnite multiplicity N.

Deﬁnition 1.3. A sequence of points {zn} in D satisfying the conditions of the above Lemma will be called a (δ, τ)-lattice on D.

Remark 1.1. The multiplicity N in the previous Lemma is independent of δ, and one can take N = 256. Moreover, if {zj} is a (δ, τ)-lattice on D, then there exists a positive δ0 2 constant (independent of δ) such that every point z in D belongs to at most N ≤ C( δ ) of the sets D(3δ0τ(zj)), where δ0 ∈ (δ, mτ ).

N T 2 Proof. Let z ∈ D(3δ0τ(zj)). Applying (1.2), that Area(D(δτ(zj)) = δ τ(zj) and j=1

δ δ D( τ(z )) ∩ D( τ(z )) = ∅ if j 6= k, 2 j 2 k we obtain

N N N X X 16 X δ Nτ(z)2 = τ(z)2 ≤ 4 τ(z )2 = Area(D( τ(z )) j δ2 2 j j=1 j=1 j=1 N 16 [ δ ≤ Area D( τ(z )) δ2 2 j j=1

N S δ Because of the inclusion D( 2 τ(zj)) ⊂ D(z, 4δ0τ(z)), we get j=1

δ 2 N ≤ C 0 , δ

where C = 256.

The next result on the density of holomorphic polynomials it is certainly well known, but for completeness we oﬀer a proof here.

Proposition 1.2. Let ω be a radial weight and 0 < p < ∞. Then the holomorphic polynomials are dense in Ap(ω).

10 p Proof. Given f ∈ A (ω) and 0 < ρ < 1, let fρ be the dilated function deﬁned by fρ(z) = − f(ρz), z ∈ D. First we want to prove that kf − fρkAp(ω) → 0 as ρ → 1 . For 0 ≤ r < 1, the classical integral means of f are deﬁned by Z 2π 1/p 1 iθ p Mp(f, r) := |f(re )| dθ , 0 < p < ∞. 2π 0

Since 0 < ρ < 1 and Mp(f, r) is increasing function of r ∈ [0, 1) (see [76, Corollary 4.21]) we get Mp(fρ, r) = Mp(f, ρr) ≤ Mp(f, r). Therefore, p p p p p+1 p Mp (f − fρ, r) ≤ 2 Mp (f, r) + Mp (fρ, r) ≤ 2 Mp (f, r). p p But the hypothesis that f ∈ A (ω) is equivalent to saying that Mp (f, r) is integrable over the interval [0, 1) with respect to the measure ω(r)rdr, and it is clear that fρ(z) → f(z) − p uniformly on compact subsets on D as ρ → 1 , which implies that Mp (f − fρ, r) → 0 for each r ∈ [0, 1). Thus by the Lebesgue dominated convergence theorem, we conclude that Z 1 p p kf − fρkAp(ω) = 2 Mp (f − fρ, r)ω(r) rdr → 0 0 − as ρ → 1 . Since fρ is analytic on D, using Runge’s theorem there is a polynomial P such that |fρ(z) − P (z)| < ε/2, for all z ∈ D and ε > 0. Finally,

kf − P kAp(ω) ≤ kf − fρk + kfρ − P k < ε, which completes the proof. Proposition 1.3. Let ω ∈ L∗ radial and 0 < p ≤ 2. Then the set E of finite linear combinations of reproducing kernels is dense in Ap(ωp/2). Proof. Suppose first that p = 2. Since E is a linear subspace of A2(ω), it is enough to 2 prove that g ≡ 0 if g ∈ A (ω) satisfies hf, giω = 0 for each f in E. But, taking f = Kz, for each z ∈ D we get g(z) = hg, Kziω = 0. This finishes the proof for p = 2. 2 p p/2 If 0 < p < 2, then A (ω) ⊂ A (ω ) continuously, that is, kfkAp(ωp/2) . kfkA2(ω). Thus, for a polynomial f and points a1, . . . , an, we have n n X X f − αkKak Ap(ωp/2) . f − αkKak A2(ω). k=1 k=1 Now, by the case p = 2 and the density of the polynomials in Ap(ωp/2), the result follows.

For the case p > 2, one can obtain the density of the span of reproducing kernels ∗ 2 Kz corresponding to some associated Bergman space A (ω∗) with a weight of the form α 2 p p/2 ω∗(z) = ω(z)τ(z) via the embedding A (ω∗) ⊂ A (ω ). For the case of the exponential weight, we will see in Chapter 4 that we can use reproducing kernels of A2(ω).

11 1.2 The class of weights W and examples

In this section, we introduce the class of weights W for which we are going to study the corresponding weighted Bergman spaces Deﬁnition 1.4. The class W consists of those radial decreasing weights of the form ω(z) = −1/2 e−2ϕ(z), where ϕ ∈ C2(D) is a radial function such that (∆ϕ(z)) τ(z), for a radial − 0 positive function τ(z) that decreases to 0 as |z| → 1 , and limr→1− τ (r) = 0. Furthermore, we shall also suppose that either there exist a constant C > 0 such that τ(r)(1 − r)−C increases for r close to 1 or 1 lim τ 0(r) log = 0. r→1− τ(r) The class W is the same class of weights considered in [9], [51] and [23], and it is straightforward to see that W ⊂ L∗. Next we give several examples of weights in the class W (these examples already appears in [51], but we give some details for completeness).

Example 1: The exponential type weights −A ω (z) = (1 − |z|2)γ exp , γ ≥ 0, σ > 0, A > 0, γ,σ (1 − |z|2)σ are in the class W with associated subharmonic function

2 2 −σ ϕγ,σ(z) = −γ log(1 − |z| ) + c (1 − |z| ) . We have that −1 2 2 2+σ ∆ϕγ,σ(z) τ(z) = (1 − |z| ) , and it is easy to see that τ(z) satisﬁes the conditions in the deﬁnition of the class W.

Example 2: For α > 1 and A > 0 the weights e α ω(r) = exp −A log , 1 − r

α e with associated subharmonic function ϕ(z) = A log 1−|z| , belong to the class W. In- deed, it is easy to see that

e α−1 ∆ϕ(z) (1 − |z|)−2 log , 1 − |z|

1−α e 2 so τ(z) = (1 − |z|) log 1−|z| , and since α > 1

−α+1 e 2 τ 0(r) log , r → 1−, 1 − r

12 0 −2 which implies that limr→1− τ(r) = limr→1− τ (r) = 0. Moreover, the function τ(r)(1 − r) increases for r close to 1. This proves that ω ∈ W.

Example 3: For α, β, γ > 0, the double exponential weight

β ω(r) = exp −γ exp (1 − r)α

β belongs to W. Indeed, the associated subharmonic function is ϕ(z) = γ exp (1−|z|)α , and a straightforward computation gives

β ∆ϕ(z) (1 − |z|)−2α−2 exp . (1 − |z|)α

α+1 −β/2 Then we can take τ(z) = (1 − |z|) exp (1−|z|)α . Since

−β/2 τ 0(r) exp , r → 1−, (1 − r)α

0 0 1 we obtain limr→1− τ(r) = limr→1− τ (r) = 0. Also, it is easy to see that limr→1− τ (r) log τ(r) = 0. This proves that ω ∈ W.

1.3 Test functions and some estimates

Since the norm of the point evaluation functional equals the norm of the reproducing kernel 2 in A (ω), the result of Lemma A also gives an upper bound for kKzkA2(ω). In order to see that this upper bound is the corresponding growth of the reproducing kernel, one needs an appropriate family of test functions, a family that can also be used in other problems such as to characterize the Carleson type measures for large weighted Bergman spaces. The following result on test functions was obtained in [51].

Lemma C. Assume that 0 < p < ∞, N ∈ N \{0} with Np ≥ 1 and ω ∈ W. Then, for each a ∈ D, there is a function Fa,N,p analytic in D with

p |Fa,N,p(z)| ω(z) 1 if |z − a| < τ(a), (1.8)

and !3N min τ(a), τ(z) |F (z)| ω(z)1/p min 1, , z ∈ . (1.9) a,N,p . |z − a| D

p Moreover, the function Fa,N,p belongs to A (ω) with

2/p kFa,N,pkAp(ω) τ(a) .

13 As a consequence we have the following estimate for the norm of the reproducing kernel, result that can be found in [9], [51]. Lemma D. Let ω ∈ W. Then

2 1 kK k 2 ω(z) , z ∈ . (1.10) z A (ω) τ(z)2 D Next result is an estimate of the reproducing kernel function for points close to the diagonal. Despite that this result is stated in [35, Lemma 3.6] we oﬀer here a proof based on Lemma 1.1, since the conditions on the weights are slightly diﬀerent.

Lemma E. Let ω ∈ W and z ∈ D. For ε > 0 suﬃciently small we have

|Kz(ξ)| kKzkA2(ω) · kKξkA2(ω), (1.11) if ξ ∈ D(ετ(z)). Proof. The upper estimate is trivial from Cauchy-Schwarz. To prove the other inequality, let ε ∈ (0, mτ ) be suﬃciently small and z ∈ D be ﬁxed such that ξ ∈ D(ετ(z)). We have 1/2 1/2 1/2 1/2 1/2 |Kz(z)|ω(z) = |Kz(z)|ω(z) − |Kz(ξ)|ω(ξ) ω(z) + |Kz(ξ)|ω(ξ) ω(z) . (1.12)

Consider 1/2 1/2 1/2 I(z, ξ) := |Kz(z)| ω(z) − |Kz(ξ)| ω(ξ) ω(z) .

By Lemma 1.1 with p = 2, we can estimate the ﬁrst term on the righthand side as follows: by Cauchy’s estimates there exists s ∈ [z, ξ] such that 1/2 1/2 I(z, ξ) ≤ ∇ |Kz|ω (s) |z − ξ| ω(z)

1/2 Z 1/2 εC(δ0) ω(z) τ(z) 2 ≤ 2 |Kz(t)| ω(t) dA(t) , τ(s) D(δ0τ(s)) for δ0 ∈ (0, mτ ) ﬁxed. Using τ(s) τ(z) and Lemma D we get

1/2 Z 1/2 ε C(δ0) ω(z) 2 I(z, ξ) ≤ |Kz(t)| ω(t) dA(t) τ(z) D(δ0τ(z)) 1/2 ε C(δ0) ω(z) ≤ kK k 2 τ(z) z A (ω) 2 ≤ ε C(δ0) ω(z) kKzkA2(ω) = ε C(δ0) ω(z) |Kz(z)|.

Putting this into (1.12) and taking ε small enough such that ε C(δ0) < 1/2, we obtain

1/2 1/2 |Kz(z)| ω(z) ≤ 2|Kz(ξ)| ω(ξ) ω(z) .

14 Therefore, again using Lemma D and the fact that τ(z) τ(ξ) for ξ ∈ D(ετ(z)), we have

kKzkA2(ω) · kKξkA2(ω) ≤ C |Kz(ξ)| obtaining the desired result.

Estimates for the ∂-equation. We recall ﬁrst the classical H¨ormander’stheorem [28] on L2-estimates for solutions of the ∂-equation.

Theorem A (H¨ormander). Let ϕ ∈ C2(D) with ∆ϕ > 0 on D. Then there exists a solution u of the equation ∂u = f such that Z Z |f(z)|2 |u(z)|2e−2ϕ(z)dA(z) ≤ e−2ϕ(z)dA(z), D D ∆ϕ(z) provided the right hand side integral is ﬁnite.

In some situations, one can use the associated weights ω∗ given by

α ω∗(z) = ω(z) τ(z) , α ∈ R. (1.13) The following result, that provides more estimates on the solutions of the ∂-equation, can be of independent interest.

α Theorem 1.4. Let ω ∈ W, and consider the associated weight ω∗(z) := ω(z) τ(z) , z ∈ D and α ∈ R. Then there exists a solution u of the equation ∂u = f such that Z Z p p/2 p p/2 p |u(z)| ω∗(z) dA(z) ≤ C |f(z)| ω∗(z) τ(z) dA(z) D D for all 1 ≤ p < ∞, provided the right hand side integral is ﬁnite. Moreover, one also has the L∞-estimate 1/2 1/2 sup |u(z)| ω∗(z) ≤ C sup |f(z)| ω∗(z) τ(z). z∈D z∈D Proof. We follow the method used in [12] where the case α = 0 was proved. By Lemma C, there are holomorphic functions Fa and some δ0 ∈ (0, mτ ) such that

−1/2 (i) |Fa(ξ)| ω(ξ) , ξ ∈ D(δ0τ(a)). min τ(ξ), τ(a)M (1.14) (ii) |F (ξ)| ≤ Cω(ξ)−1/2 , (a, ξ) ∈ × , a |a − ξ| D D

Let δ1 < δ0. Then, there is a sequence {zn}n≥1 such that {D(δ1τ(zn))} is a covering of D of ﬁnite multiplicity N and satisfy the other statements of Lemma B. Let χn be a partition of unity subordinate to the covering D(δ1τ(zn)). Consider Z f(ξ) χn(ξ) Snf(z) = Fzn (z) dA(ξ). D (ξ − z) Fzn (ξ) 15 Since Fzn are holomorphic functions on D, by the Cauchy-Pompieu formula, we have

∂Snf(z) = f(z) χn(z), n = 1, 2....

Then, ∞ Z X −1/2 1/2 Sf(z) = Snf(z) = ω∗(z) G(z, ξ) f(ξ) ω∗(ξ) dA(ξ), n=1 D where ∞ X Fz (z) χn(ξ) G(z, ξ) = n ω (ξ)−1/2 ω (z)1/2. ξ − z F (ξ) ∗ ∗ n=1 zn

Since χn is a partition of the unity, we have

∞ ∞ X X ∂(Sf) = ∂(Snf) = f χn = f n=1 n=1 on D, so that Sf solves the equation ∂Sf(z) = f(z).

Now we are going to prove that

Z dA(ξ) |G(z, ξ)| . 1. (1.15) D τ(ξ)

First we consider the covering of {ξ ∈ D : |z − ξ| > δ2τ(z)} given by

k−1 k Rk(z) = {ξ ∈ D : 2 δ2τ(z) < |z − ξ| ≤ 2 δ2τ(z)}, k = 0, 1, 2,...

δ0 Let 4 δ1 < δ2 < 5 and z ∈ D be ﬁxed. If ξ ∈ D(δ2τ(z)) ∩ D(δ1τ(zn)), using (1.2)

|z − zn| ≤ |z − ξ| + |ξ − zn| ≤ δ2τ(z) + δ1τ(zn)

≤ 4δ2τ(zn) + δ1τ(zn) < δ0τ(zn), that implies z ∈ D(δ0τ(zn)). Using (1.2) and property (i) of (1.14), it follows

−1/2 −1/2 1/2 ∞ ω(z) ω∗(ξ) ω∗(z) X 1 |G(z, ξ)| χ (ξ) . . |ξ − z| ω(ξ)−1/2 n . |ξ − z| n=1 Therefore, using (1.2) and polar coordinates, we get

Z dA(ξ) 1 Z 1 |G(z, ξ)| . dA(ξ) . 1. (1.16) D(δ2τ(z)) τ(ξ) τ(z) D(δ2τ(z)) |ξ − z|

16 If ξ ∈ D r D(δ2τ(z)) ∩ D(δ1τ(zn)), we show that z∈ / D(δ1τ(zn)). In fact, if not, z ∈ D(δ1τ(zn)), then using (1.2) we have

|z − zn| > |z − ξ| − |ξ − zn| > δ2τ(z) − δ1τ(zn) ≥ (δ2/2 − δ1)τ(zn) ≥ δ1τ(zn), this implies a contradiction with our assumption. Thus,

|z − ξ| ≤ |z − zn| + |zn − ξ| ≤ |z − zn| + δ1τ(zn)

≤ 2|z − zn|.

Also, using τ(ξ) τ(zn) we get

|z − zn| |z − ξ| ≥ C . min τ(z), τ(zn) min τ(z), τ(ξ)

Then, again using τ(ξ) τ(zn) and property (ii) of (1.14) with

M > max(1; −α/2; 1 + α/2), we have

∞ X |Fz (z)| χn(ξ) |G(z, ξ)| ≤ C n ω (ξ)−1/2ω (z)1/2 |ξ − z| |F (ξ)| ∗ ∗ n=1 zn

−1/2 −1/2 1/2 M ∞ ω(z) ω∗(ξ) ω∗(z) min τ(z), τ(ξ) X ≤ C χ (ξ) (1.17) |ξ − z| ω(ξ)−1/2 |ξ − z| n n=1

α M τ(z) 2 min τ(z), τ(ξ) ≤ C α . τ(ξ) 2 |ξ − z| |ξ − z|

Then,

Z dA(ξ) |G(z, ξ)| τ(ξ) DrD(δ2τ(z)) −(2+α) M Z τ(ξ) 2 min τ(z), τ(ξ)) ≤ Cτ(z)α/2 dA(ξ). |ξ − z| |ξ − z| DrD(δ2τ(z))

17 • If 2 + α > 0.

∞ Z dA(ξ) X Z 1 |G(z, ξ)| ≤ Cτ(z)M−1 dA(ξ) τ(ξ) |ξ − z|M+1 DrD(δ2τ(z)) k=1 Rk(z)

∞ X Z 1 ≤ Cτ(z)M−1 dA(ξ) k M+1 k=1 Rk(z) 2 τ(z)

∞ X 1 ≤ C 1. 2k(M−1) . k=1 • If 2 + α ≤ 0. Using the condition (B) in the deﬁnition of the class L, it follows that

k τ(ξ) ≤ C2 δ2τ(z), ξ ∈ Rk(z) k = 0, 1, 2,...

So,

Z dA(ξ) Z τ(ξ)−(2+α)/2τ(z)M |G(z, ξ)| ≤ Cτ(z)α/2 dA(ξ) τ(ξ) |ξ − z|M+1 DrD(δ2τ(z)) DrD(δ2τ(z))

∞ k −(2+α)/2 X Z 2 τ(z) ≤ Cτ(z)M+α/2 dA(ξ) k M+1 k=1 Rk(z) 2 τ(z)

∞ X 1 ≤ C 1. 2k(M+α/2) . k=1 This together with (1.16) establish (1.15).

Using (1.15), it is straightforward that the L∞-estimate holds. Our next goal is to prove the inequality Z Z p p/2 p p/2 p |Sf(z)| ω∗(z) dA(z) . |f(z)| ω∗(z) τ(z) dA(z). D D

1/2 R Consider g(ξ) := f(ξ) ω∗(ξ) and T g(z) := G(z, ξ) g(ξ)dA(ξ). Then, the last inequality D translates Z Z p p p |T g(z)| dA(z) . |g(z)| τ(z) dA(z). D D

18 Therefore, using H¨older’sinequality and (1.15), we have Z p |T g(z)|p ≤ |G(z, ξ)| |g(ξ)| dA(ξ) D

Z Z dA(ξ)p−1 ≤ |g(ξ)|p τ(ξ)p−1|G(z, ξ)| dA(ξ) |G(z, ξ)| D D τ(ξ)

Z p p−1 . |g(ξ)| τ(ξ) |G(z, ξ)| dA(ξ). D These and Fubini’s theorem give Z Z Z p p p−1 |T g(z)| dA(z) . |g(ξ)| τ(ξ) |G(z, ξ)| dA(ξ) dA(z) D D D

Z Z p p−1 . |g(ξ)| τ(ξ) |G(z, ξ)| dA(z) dA(ξ). D D

Now, using the expression of the kernel G(z, ξ) and the fact that χn are supported in D(δ1τ(zn)), we obtain

Z Z Z ∞ 1/2 X |Fz (z)| χn(ξ) ω∗(z) |T g(z)|pdA(z) |g(ξ)|p τ(ξ)p−1 n dA(z) dA(ξ) . |ξ − z| |F (ξ)| ω (ξ)1/2 D D D n=1 zn ∗

∞ Z −1/2 Z X ω∗(ξ) |Fz (z)| |g(ξ)|p τ(ξ)p−1 n ω (z)1/2 dA(z) χ (ξ) dA(ξ) . |F (ξ)| |ξ − z| ∗ n n=1 D zn D

∞ Z −1/2 Z X ω∗(ξ) |Fz (z)| |g(ξ)|p τ(ξ)p−1 n ω (z)1/2 dA(z) dA(ξ). . |F (ξ)| |ξ − z| ∗ n=1 D(δ1τ(zn)) zn D

−1/2 By using |Fzn (ξ)| ω(ξ) , ξ ∈ D(δ1τ(zn)), it follows that Z |T g(z)|pdA(z) D ∞ Z Z (1.18) X |Fz (z)| |g(ξ)|p τ(ξ)p−1−α/2 n ω (z)1/2 dA(z) dA(ξ). . |ξ − z| ∗ n=1 D(δ1τ(zn)) D We claim that Z |Fzn (z)| 1/2 1+α/2 ω∗(z) dA(z) . τ(ξ) , ξ ∈ D(δ1τ(zn)). (1.19) D |ξ − z| 19 Assuming the claim, we can ﬁnish the proof. Indeed, since {D(δ1τ(zn))} is a covering of D of ﬁnite multiplicity, using the claim we have Z ∞ Z p X p p |T g(z)| dA(z) . |g(ξ)| τ(ξ) dA(ξ) D n=1 D(δ1τ(zn))

Z p p . |g(ξ)| τ(ξ) dA(ξ). D Therefore it remains to prove the inequality (1.19). We split this integral in two parts: one integrating over the disk D(δ2τ(ξ)) and the other one over D r D(δ2τ(ξ)). We compute the ﬁrst integral, using (i) of (1.14), τ(ξ) τ(zn) τ(z), z ∈ D(δ2τ(ξ)) and by using polar coordinates, we obtain Z Z |Fzn (z)| 1/2 α/2 dA(z) 1+α/2 ω∗(z) dA(z) . τ(ξ) . τ(ξ) . (1.20) D(δ2τ(ξ)) |ξ − z| D(δ2τ(ξ)) |z − ξ| Now we consider Z |F (z)| I(ξ) := zn ω (z)1/2 dA(z), ξ ∈ D(δ τ(z )). |ξ − z| ∗ 1 n DrD(δ2τ(ξ))

For z∈ / D(δ2τ(ξ))

|z − ξ| ≤ |z − zn| + |zn − ξ| ≤ |z − zn| + δ1τ(zn)

2δ1 2δ1 ≤ |z − zn| + |z − zn| ≤ 1 + |z − zn|. δ2 δ2

Then, again by using τ(ξ) τ(zn), we obtain

min τ(z ), τ(z) min τ(ξ), τ(z) n ≤ C |z − zn| |z − ξ| This together with (1.14) taking M > max(1, 2 + α/2) give

M Z τ(z)α/2 min τ(zn), τ(z) I(ξ) dA(z) . |ξ − z| |z − z |M DrD(δ2τ(ξ)) n M Z min τ(ξ), τ(z) τ(z)α/2 dA(z). . |z − ξ|M+1 DrD(δ2τ(ξ))

20 k • Suppose ﬁrst that α ≥ 0. By τ(z) ≤ C2 τ(ξ), for every z ∈ Rk(ξ), k = 0, 1, 2,..., we get

∞ k α/2 M X Z 2 τ(ξ) τ(ξ) I(ξ) dA(z) . k M+1 k=1 Rk(ξ) 2 τ(ξ) ∞ X 1 τ(ξ)1+α/2 τ(ξ)1+α/2. . 2k(M−2−α/2) . k=1 • If α < 0, then M Z min τ(ξ), τ(z) I(ξ) τ(z)α/2 dA(z) . |z − ξ|M+1 DrD(δ2τ(ξ))

∞ X Z τ(ξ)M+α/2 dA(z) . |z − ξ|M+1 k=1 Rk(ξ)

∞ X 1 τ(ξ)1+α/2 τ(ξ)1+α/2. . 2k(M−1) . k=1 This together with (1.20) establish (1.19). The proof is complete.

1.4 Carleson type measures

Now we are going to recall some results on the boundedness of the embedding i : X → Lq(µ) for some spaces X of analytic functions. We need ﬁrst the notion of an s-Carleson measure. Given an arc I of the unit circle T := ∂D, the Carleson box associated to I is |I| z S(I) = z ∈ : 1 − < |z| < 1, ∈ I . D 2π |z|

For α > 0, a positive Borel measure µ on D is called an α-Carleson measure if µS(I) kµkCMα := sup α < ∞. I⊂T |I| The α-Carleson measures are the appropriate geometric object in order to describe the boundedness of i : Hp → Lq(µ), as the following classical result of Carleson [10, 11] (case q = p) and Duren [20] (case p < q) shows. Theorem B (Carleson-Duren). Let 0 < p ≤ q < ∞. Then the embedding i : Hp → Lq(µ) is bounded if and only if µ is an q/p-Carleson measure.

21 The corresponding result for 0 < q < p < ∞ it seems that was considered a folklore theorem for a long time. It can be found stated in [65]. A detailed proof can be found in [40] or [69].

Theorem C. Let 0 < q < p < ∞. The following conditions are equivalent:

(a) i : Hp → Lq(µ) is bounded.

p R dµ(z) p−q (b) The function Aµ1(ζ) = Γ(ζ) 1−|z| is in L (T).

p (c) The sweep µe of µ belongs to L p−q (T). We recall that, for a measure µ on D, the sweep of µ onto T is deﬁned as 1 Z 1 − |z|2 µ(ζ) = dµ(z). e 2 2π D |ζ − z|

A description of q-Carleson measures for Ap(ωp/2), 0 < p, q < ∞, for weights ω in the class W was obtained in [51]. A positive measure µ is q-Carleson for Ap(ωp/2) when the p p/2 q inclusion Iµ : A (ω ) −→ L (D, dµ) is bounded. Next results were proved in [51, Theorem 1].

Theorem D. Let ω ∈ W and µ be a ﬁnite positive Borel measure on D. Let 0 0 we have Z 1 −q/2 Kµ,ω := sup 2q/p ω(ξ) dµ(ξ) < ∞. (1.21) a∈D τ(a) D(δτ(a)) q Moreover, in that case, Kµ,ω kIµk . Ap(ωp/2)→Lq(D,dµ)

Theorem E. Let ω ∈ W and µ be a ﬁnite positive Borel measure on D. Let 0 0 we have 1 Z lim sup ω(ξ)−q/2 dµ(ξ) = 0. (1.22) − 2q/p r→1 |a|>r τ(a) D(δτ(a))

Theorem F. Let ω ∈ W and let µ be a ﬁnite positive Borel measure on D. Let 0 < q < p < ∞. The following conditions are equivalent:

p p/2 q (a) Iµ : A (ω ) −→ L (D, dµ) is compact.

22 p p/2 q (b) Iµ : A (ω ) −→ L (D, dµ) is bounded. (c) For each suﬃciently small δ > 0, the function Z 1 −q/2 Fµ(z) = 2 ω(ξ) dµ(ξ) τ(z) D(δτ(z))

p belongs to L p−q (D, dA). Moreover, one has q p kIµk p p/2 q kFµk . A (ω )→L (µ) L p−q (D) 1.5 Atomic Decomposition

In this Section we are going to obtain an atomic decomposition for the large weighted Bergman space A2(ω), that is, we show that every function in the Bergman spaces A2(ω) with ω in the class W can be decomposed into a series of kernels functions. Atomic decomposition for standard Bergman spaces were obtained by Coifman and Rochberg [15] and has become a powerful tool in the study of such spaces. We refer to the books [53, 77, 76] for a modern proof of these results.

Proposition 1.5. Let ω ∈ W, and {zk}k∈N ⊂ D be the sequence that it is deﬁned in Lemma B. The function given by

∞ X 1/2 F (z) := λk ω(zk) τ(zk) Kzk (z) k=0

2 2 belongs to A (ω) for every sequence λ = {λk} ∈ ` . Moreover,

kF kA2(ω) . kλk`2 . PN Proof. First we show that the partial sums FN = k=0 fk converges uniformly on compact subsets of D which proves that F deﬁnes an analytic function on D. Using Cauchy-Schwarz inequality, Lemma A and the norm estimate of kKzkA2(ω), we have

N 1/2 N 1/2 X 2 X 2 2 |FN (z)| ≤ |λk| τ(zk) |Kzk (z)| ω(zk) k=0 k=0 N 1/2 N Z 1/2 X 2 X 2 . |λk| |Kz(ξ)| ω(ξ) dA(ξ) k=0 k=0 D(δτ(zk)) ω(z)−1/2 kλk 2 kK k 2 kλk 2 , . ` z A (ω) . τ(z) ` which proves that the series deﬁning F converges uniformly on compact subsets of D.

23 2 To prove that F ∈ A (ω) with kF kA2(ω) . kλk`2 , by duality we must show that 2 |hF, giω| . kλk`2 · kgkA2(ω) for g ∈ A (ω). Clearly,

X 1/2 hF, giω ≤ |λk| ω(zk) τ(zk) |g(zk)| k !1/2 X 2 2 ≤ kλk`2 |g(zk)| ω(zk) τ(zk) . k

Finally, by Lemma , and since {zn} is a τ-lattice, Z X 2 2 X 2 2 |g(zk)| ω(zk) τ(zk) . |g(ζ)| ω(z) dA(z) . kgkA2(ω). k k D(δτ(zk)) This ﬁnishes the proof.

Lemma 1.6. Let ω ∈ E. There is a sequence {zn} ⊂ D such that ∞ X p p/2 2 p |f(zn)| ω(zn) τ(zn) & kfkAp(ωp/2), n=0 for all f ∈ Ap(ωp/2) and p > 0.

Proof. Let {zn} be an (ε, τ)-lattice on D (that exists by Lemma B) with ε > 0 small enough to be speciﬁed later. Let f ∈ Ap(ωp/2). We consider

∞ X p p/2 2 If (n) := |f(zn)| ω(zn) τ(zn) . n=0 We have Z p p p/2 kfkAp(ωp/2) = |f(z)| ω(z) dA(z) D

∞ Z p X 1/2 1/2 2 ≤ C |f(z)|ω(z) − |f(zn)|ω(zn) dA(z) + Cε If (n) . n=0 D(ετ(zn))

For z ∈ D(ετ(zn)), there exists ξn,z ∈ [z, zn] such that p 1/2 1/2 1/2 p p |f(z)|ω(z) − |f(zn)|ω(zn) ≤ ∇(|f| ω )(ξn,z) |z − zn|

p p 1/2 p ≤ ε τ(zn) ∇(|f| ω )(ξn,z) .

24 This together with Lemma 1.1, with δ0 ∈ (0, mτ ) ﬁxed, yields Z p 1/2 1/2 |f(z)|ω(z) − |f(zn)|ω(zn) dA(z) D(ετ(zn))

Z Z ! p p 1 p p/2 ≤ C ε τ(zn) p+2 |f(η)| ω(η) dA(η) dA(z). D(ετ(zn)) τ(ξn,z) D(δ0τ(ξn,z))

Using that τ(ξn,z) τ(zn) and D(δ0τ(ξn,z)) ⊂ D(3δ0τ(zn)) for z ∈ D(ετ(zn)), we obtain Z p 1/2 1/2 |f(z)|ω(z) − |f(zn)|ω(zn) dA(z) D(ετ(zn))

Z ≤ C εp+2 |f(η)|pω(η)p/2dA(η) . D(3δ0τ(zn)) Therefore, ∞ Z p p+2 X p p/2 2 kfkAp(ωp/2) ≤ C ε |f(η)| ω(η) dA(η) + Cε If (n). n=0 D(3δ0τ(zn))

−2 By Lemma B every point z ∈ D belongs to at most Cε of the sets D(3δ0τ(zn)), and therefore

p p p 2 kfkAp(ωp/2) ≤ Cε kfkAp(ωp/2) + C ε If (n).

Thus, taking ε > 0 so that Cεp < 1/2, we get the desired result.

It is worth mentioning that, what actually Lemma 1.6 says, is that an (ε, τ)-lattice with ε > 0 small enough, is a sampling sequence for the Bergman space Ap(ωp/2). Recall that p p/2 {zn} ⊂ D is a sampling sequence for the Bergman space A (ω ) if

p X p p/2 2 kfkAp(ωp/2) |f(zn)| ω(zn) τ(zn) n for any f ∈ Ap(ωp/2). Just note that Lemma 1.6 gives one inequality, and the other follows by standard methods using Lemma A and the lattice properties. Sampling sequences on the classical Bergman space were characterized by K. Seip [59] (see also the monographs [22] and [60]). For sampling sequences on large weighted Bergman spaces we refer to [9].

25 Now we are ready to prove the result on the atomic decomposition of large weighted 2 Bergman spaces A (ω). We use the notation kz for the normalized reproducing kernels in A2(ω), that is Kz kz = . kKzkA2(ω)

Theorem 1.7. Let ω ∈ W. There exists a lattice {zn} ⊂ D such that:

2 (i) For any λ = {λn} ∈ ` , the function X f(z) = λn kzn (z) n

2 is in A (ω) with kfkA2(ω) ≤ Ckλk`2 .

2 2 (ii) For every f ∈ A (ω) exists λ = {λn} ∈ ` such that X f(z) = λn kzn (z) n

and kλk`2 ≤ C kfkA2(ω).

−1 1/2 Proof. Because of the norm estimate kKzn kA2(ω) ω(zn) τ(zn), part (i) is just Propo- sition 1.5. In order to prove (ii), we deﬁne a linear operator S : `2 −→ A2(ω) given by ∞ X S({λn}) := λn kzn . n=0 By (i), the operator S is bounded. The adjoint operator S∗ : A2(ω) → `2 is deﬁned by

∗ X ∗ hSx, fiω = hx, S fi` = xn (S f)n. n

2 2 ∗ for every x ∈ ` and f ∈ A (ω). To compute S , let en denote the vector that equals 1 at the n-th coordinate and equals 0 at the other coordinates. Then Sen = kzn , and using the reproducing formula we get

∗ ∗ (S f)n = hen,S fi` = hSen, fiω = hkzn , fiω

−1 f(zn) = kKzn kA2(ω) · hKzn , fiω = . 2 kKzn kA (ω)

Hence, S∗ : A2(ω) −→ `2 is given by ∗ ∗ f(zn) S f = {(S f)n} = . 2 kKzn kA (ω) n

26 We must prove that S is surjective in order to ﬁnish the proof of this case. By a classical result in functional analysis, it is enough to show that S∗ is bounded below. By Lemma 1.6 and the estimate for the norm of Kz we obtain

∞ ∗ 2 X 2 2 2 kS fk`2 |f(zn)| ω(zn) τ(zn) & kfkA2(ω), n=0 which shows that S∗ is bounded below. Finally, once the surjectivity is proved, the estimate kλk`2 ≤ C kfkA2(ω) is an standard application of the open mapping theorem. The proof is complete.

27 28 Chapter 2

Toeplitz operators

Throughout this chapter, µ will denote a ﬁnite positive Borel measure on D and ω would be a weight in the class W so that we can apply all the results of the previous chapter. The purpose of this chapter is to characterize the boundedness, compactness and membership in Schatten ideals of Toeplitz operators acting on A2(ω).

2 ω Deﬁnition 2.1. Let Kz be the reproducing kernel of A (ω). The Toeplitz operator Tµ with symbol µ is given by Z ω Tµf(z) = Tµ f(z) := f(ξ) Kz(ξ) ω(ξ) dµ(ξ). D

Note that Tµ is very loosely defined here, because it is not clear when the integrals above will converge, even if the measure µ is finite. We suppose that µ is a finite positive Borel measure that satisfies the condition Z 2 |Kz(ξ)| ω(ξ) dµ(ξ) < ∞. (2.1) D

2 Then, the Toeplitz operator Tµ is well-defined on a dense subset of A (ω). In fact, by Proposition 1.3, the set E of finite linear combinations of reproducing kernels is dense in A2(ω). Therefore, it follows from condition (2.1) and the Cauchy-Schwartz inequality that Tµ(f) is well defined for any f ∈ E.

There is a lot of work on the study of Toeplitz operators acting on several spaces of holomorphic functions [13, 29, 39, 43, 57, 72], and the theory is especially well understood in the case of Hardy spaces or standard Bergman spaces (see [77] and the references therein). Luecking [39] was the ﬁrst to study Toeplitz operators on the Bergman spaces with measures as symbols, and the study of Toeplitz operators acting on large weighted Bergman spaces was initiated by Lin and Rochberg [35]

29 2.1 Boundedness and compactness

In this section we describe the boundedness of Toeplitz operators on large Bergman spaces. For δ ∈ (0, mτ ), we deﬁne a new function µbδ (the averaging function of µ) on D by µ(D(δτ(z)) µ (z) := , z ∈ . bδ τ(z)2 D

Theorem 2.1. Let ω ∈ W. Then

2 2 (i) Tµ : A (ω) → A (ω) is bounded if and only if for each δ > 0 small enough, one has

Cµ := sup µbδ(z) < ∞. (2.2) z∈D

Moreover, in that case, kTµk Cµ.

2 2 (ii) Tµ : A (ω) → A (ω) is compact if and only if for each δ > 0 small enough, one has

lim sup µδ(z) = 0. (2.3) − b r→1 |z|>r

For the proof, we need ﬁrst the following lemma.

∞ Lemma 2.2. Let ω ∈ W, and assume that µbδ is in L (D) for some small δ > 0. Then Z 2 hTµf, giω = f(ζ) g(ζ) ω(ζ) dµ(ζ), f, g ∈ A (ω). D Proof. Being ω a radial weight, the polynomials are dense in A2(ω) and we may assume that g is an holomorphic polynomial. Because of Theorem D, the condition implies that dν = ω dµ is a Carleson measure for A2(ω). Then Z Z |f(ζ)| |Kz(ζ)| ω(ζ) dµ(ζ) |g(z)| ω(z) dA(z) D D Z ≤ kfkL2(ν) kKzkL2(ν) |g(z)| ω(z) dA(z) D Z . kfkA2(ω) · kgk∞ kKzkA2(ω) ω(z) dA(z) D Z ω(z)1/2 . kfkA2(ω) · kgk∞ dA(z), D τ(z)

30 and this is ﬁnite (see [51, Lemma 2.3] for example). Thus, Fubini’s theorem gives Z Z hTµf, giω = f(ζ) Kz(ζ) ω(ζ) dµ(ζ) g(z) ω(z) dA(z) D D

Z Z = f(ζ) g(z) Kζ (z) ω(z) dA(z) ω(ζ) dµ(ζ) D D

Z Z = f(ζ) hg, Kζ iω ω(ζ) dµ(ζ) = f(ζ) g(ζ) ω(ζ) dµ(ζ). D D

Proof of Theorem 2.1: boundedness

2 Assume ﬁrst that Tµ is bounded on A (ω). For ﬁxed a ∈ D, one has Z 2 TµKa(a) = |Ka(z)| ω(z) dµ(z). D

By Lemma E, there is δ ∈ (0, mτ ) such that |Ka(z)| kKzkA2(ω) · kKakA2(ω), for every z ∈ D(δτ(a)). This together with the norm estimate given in Lemma D and the fact that τ(z) τ(a) for z ∈ D(δτ(a)) gives Z 2 TµKa(a) ≥ |Ka(z)| ω(z) dµ(z) D(δτ(a))

Z 2 2 & kKzkA2(ω) kKakA2(ω) ω(z) dµ(z) D(δτ(a))

µ(D(δτ(a)) µ (a) = bδ . ω(a) τ(a)4 ω(a) τ(a)2 Therefore, by Lemma A and the estimate of the norm of the reproducing kernels, we obtain

2 1/2 µbδ(a) . ω(a) τ(a) |TµKa(a)| ≤ ω(a) τ(a) kTµKakA2(ω) (2.4) 1/2 ≤ ω(a) τ(a) kTµk · kKakA2(ω) . kTµk.

Conversely, suppose that (2.2) holds. Let f, g ∈ A2(ω). By Lemma 2.2 and because

31 dν = ωdµ is a Carleson measure for A2(ω) with kI k2 C := sup µ (z) (see Theorem ν µ z∈D bδ D), we have Z

hTµf, giω ≤ |f(z)| |g(z)| dν(z) ≤ kfkL2(ν) · kgkL2(ν) . CµkfkA2(ω) · kgkA2(ω). D

2 This shows that Tµ is bounded on A (ω) with kTµk . Cµ ﬁnishing the proof.

Proof of Theorem 2.1: compactness

2 Let kz be the normalized reproducing kernels in A (ω). From (2.4) in the proof of the boundedness part, and the estimate for kKzkA2(ω), we have

µbδ(z) . kTµkzkA2(ω).

− From Lemma A, it is easy to see that kz converges to zero weakly as |z| → 1 . Thus, if Tµ is compact, from Theorem 1.14 in [77], we obtain (2.3).

2 Conversely, suppose (2.3) holds, and let {fn} be a sequence in A (ω) converging to zero weakly. To prove compactness, we must show that kTµfnkA2(ω) → 0. By the proof of the boundedness, we have kTµfnkA2(ω) . kfnkL2(ν), 2 2 with dν := ωdµ. By Theorem E, our assumption (2.3) implies that Iν : A (ω) −→ L (D, dν) is compact, which implies that kfnkL2(ν) tends to zero. Hence kTµfnkA2(ω) → 0 proving that Tµ is compact. The proof is complete.

2.2 Membership in Schatten classes

In this section we will provide a full characterization of the Schatten class membership of Toeplitz operators acting on large weighted Bergman spaces A2(ω). The case of the 2 classical weighted Bergman spaces Aα with α > −1 was obtained by D. Luecking [39], and can be found also in Zhu’s book [77].

2 2 Given 0 < p < ∞, let Sp(A (ω)) denote the Schatten p-class of operators from A (ω) to A2(ω), it consists of those compact operators T : A2(ω) −→ A2(ω) with its sequence of p singular numbers λn belonging to ` , the p-summable sequence space. We recall that the singular numbers of a compact operator T are the square root of the eigenvalues of the ∗ ∗ 2 positive operator T T, where T denotes the adjoint of T. One has T ∈ Sp(A (ω)) if and ∗ 2 only if T T ∈ Sp/2(A (ω)). Also, the compact operator T admits a decomposition of the form ∞ X T = λnh., eniA2(ω)en, n=1

32 2 where {λn} are the singular numbers of T and {en} is an orthonormal basis in A (ω). For 2 p ≥ 1, the class Sp(A (ω)) is Banach space equipped with the norm

∞ 1/p X p kT kSp := |λn| , n=1 while for 0 < p < 1 one has the inequality [45, Theorem 2.8]

kS + T kp ≤ kSkp + kT kp . Sp Sp Sp

Two special cases are worth mentioning: S1 is called the trace class, and S2 is called the Hilbert-Schmidt class. For a positive operator T, let tr(T ) denote the usual trace deﬁned by ∞ X tr(T ) := hT en, eniω = kT kS1 . n=1

More background information about the Schatten classes Sp can be found in [77, Chapter 1] for example.

The main result of this section is the following:

In this section, we are going to describe those positive Borel measures µ for which the 2 Toeplitz operator Tµ belongs to the the Schatten ideal Sp(A (ω)), for weights ω ∈ W. In order to obtain such a characterization, we need to introduce ﬁrst some concepts.

We deﬁne the ω-Berezin transform Bωµ of the measure µ as Z 2 Bωµ(z) := |kz(ξ)| ω(ξ) dµ(ξ), z ∈ D, D 2 where kz are the normalized reproducing kernels in A (ω). We also recall that, for δ ∈ (0, mτ ), the averaging function µbδ is given by µ(D(δτ(z)) µ (z) := , z ∈ . bδ τ(z)2 D

We also consider the measure λτ given by dA(z) dλ (z) = , z ∈ . τ τ(z)2 D Proposition 2.3. Let 1 ≤ p < ∞, and ω ∈ W. The following conditions are equivalent :

p (a) The function Bωµ is in L (D, dλτ ).

p (b) The function µbδ is in L (D, dλτ ) for any δ ∈ (0, mτ ) small enough.

33 (c) The sequence {µ (z )} is in `p for any (δ, τ)-lattice {z } with δ ∈ (0, mτ ) suﬃciently bδ n n 4 small.

Proof. (a) ⇒ (b). By Lemma E, for all δ ∈ (0, mτ ) suﬃciently small, one has

|Kz(ζ)| kKzkA2(ω) · kKζ kA2(ω), ζ ∈ D(δτ(z)) Then Z Z 2 −2 2 Bωµ(z) = |kz(ζ)| ω(ζ) dµ(ζ) ≥ kKzkA2(ω) |Kz(ζ)| ω(ζ) dµ(ζ) D D(δτ(z))

Z 2 kKζ kA2(ω) ω(ζ) dµ(ζ) µbδ(z). D(δτ(z))

p Since Bωµ is in L (D, dλτ ), this gives (b).

(b) ⇒ (c). Since µbδ(zn) . µb4δ(z), for z ∈ D(δτ(zn)), then X X Z dA(z) Z µ (z )p µ (z)p µ (z)p dλ (z). bδ n . b4δ τ(z)2 . b4δ τ n n D(δτ(zn)) D

We use Lemma A in order to obtain

Z Z Z 2 1 2 |Kz(s)| ω(s) dµ(s) . 2 |Kz(ξ)| ω(ξ) dA(ξ) dµ(s) D(δτ(zn)) D(δτ(zn)) τ(s) D(δτ(s))

Z 2 . |Kz(ξ)| ω(ξ) dA(ξ) µbδ(zn). D(3δτ(zn))

If p > 1, by H¨older’sinequality,

Z !p X 2 |Kz(s)| ω(s) dµ(s) n D(δτ(zn))

Z 2(p−1) X kK k |K (ξ)|2 ω(ξ) dA(ξ) µ (z )p. . z A2(ω) z bδ n n D(3δτ(zn))

34 This gives

Z X Z Z B µ(z)p dλ (z) µ (z )p |K (z)|2 kK k−2 dλ (z) ω(ξ) dA(ξ). ω τ . bδ n ξ z A2(ω) τ D n D(3δτ(zn)) D

2 −2 −1 Since kKzkA2(ω) τ(z) ω(z) , we have Z 2 −2 2 −2 −1 |Kξ(z)| kKzkA2(ω)dλτ (z) kKξkA2(ω) τ(ξ) ω(ξ) . D Putting this estimate in the previous inequality, we ﬁnally get Z p X p Bωµ(z) dλτ (z) . µbδ(zn) , p ≥ 1. D n This ﬁnishes the proof.

Remark: It should be observed that the equivalence of (b) and (c) in Proposition 2.3 con- tinues to hold for 0 < p < 1 as well as the implication (a) implies (b). In order to get equivalence with condition (a) even in the case 0 < p < 1, it seems that one needs Lp- integral estimates for reproducing kernels, estimates that are not available nowadays.

Next Lemma is the analogue to our setting of a well known result for standard Bergman spaces.

Lemma 2.4. Let ω ∈ W, and T be a positive operator on A2(ω). Let Te be the Berezin transform of the operator T deﬁned by

Te(z) = hT kz, kziω, z ∈ D.

p (a) Let 0 < p ≤ 1. If Te ∈ L (D, dλτ ) then T is in Sp.

p (b) Let p ≥ 1. If T is in Sp then Te ∈ L (D, dλτ ).

p Proof. Let p > 0. The positive operator T is in Sp if and only if T is in the trace class 2 p S1. Fix an orthonormal basis {ek} of A (ω). Since√ T is positive, it belongs to the trace P p p class if and only if khT ek, ekiω < ∞. Let S = T . Then

X p X 2 hT ek, ekiω = kSekkA2(ω). k k

35 Now, by Fubini’s theorem and Parseval’s identity, we have Z Z X 2 X 2 X 2 kSekkA2(ω) = |Sek(z)| ω(z) dA(z) = |hSek,Kziω| ω(z) dA(z) k k D k D

Z ! Z X 2 2 = |hek,SKziω| ω(z) dA(z) = kSKzkA2(ω) ω(z) dA(z) D k D

Z Z p p 2 = hT Kz,Kziω ω(z) dA(z) = hT kz, kziωkKzkA2(ω) ω(z) dA(z) D D

Z p hT kz, kziω dλτ (z). D

Hence, both (a) and (b) are consequences of the inequalities (see [77, Proposition 1.31])

p p p hT kz, kziω ≤ [hT kz, kziω] = [Te(z)] , 0 < p ≤ 1 and

p p p [Te(z)] = [hT kz, kziω] ≤ hT kz, kziω, p ≥ 1.

This ﬁnishes the proof of the lemma.

p Proposition 2.5. Let ω ∈ W. If 0 < p ≤ 1 and Bωµ is in L (D, dλτ ), then Tµ belongs to 2 2 p Sp(A (ω)). Conversely, if p ≥ 1 and Tµ is in Sp(A (ω)), then Bωµ ∈ L (D, dλτ ).

p 2 Proof. If Bωµ is in L (D, dλτ ), then it is easy to see that Tµ is bounded on A (ω) (just use the discrete version in Proposition 2.3 to see that the condition in Theorem 2.1 holds).

Therefore, the result is a consequence of Lemma 2.4 since Tfµ(z) = Bωµ(z).

Now we are almost ready for the characterization of Schatten class Toeplitz operators, but we need ﬁrst some technical lemmas on properties of lattices. We use the notation

|z − ζ| d (z, ζ) = , z, ζ ∈ . τ min(τ(z), τ(ζ)) D

36 Lemma 2.6. Let τ ∈ L, and {zj} be a (δ, τ)-lattice on D. For each ζ ∈ D, the set

m Dm(ζ) = z ∈ D : dτ (z, ζ) < 2 δ contains at most K points of the lattice, where K depends on the positive integer m but not on the point ζ.

Proof. Let K be the number of points of the lattice contained in Dm(ζ). Due to the Lipschitz condition (B), we have

m τ(ζ) ≤ τ(zj) + c2|ζ − zj| ≤ (1 + c22 δ) τ(zj) = Cm τ(zj). Then   X [ δ K · τ(ζ)2 ≤ C2 τ(z )2 C2 · Area D τ(z ) . m j . m  4 j  zj ∈Dm(ζ) zj ∈Dm(ζ)

As done before, we also have τ(zj) ≤ Cmτ(ζ), if zj ∈ Dm(ζ). From this we easily see that δ D τ(z ) ⊂ D c2mδτ(ζ) 4 j

δ for some constant c. Since the sets {D( 4 τ(zj))} are pairwise disjoints, we have [ δ D τ(z ) ⊂ D c2mδτ(ζ) . 4 j zj ∈Dm(ζ) Therefore, we get

2 2 m 2 2m 2 K · τ(ζ) ≤ Cm · Area D c2 δτ(ζ) . Cm2 τ(ζ) ,

that implies K ≤ C24m.

Next, we use the result just proved to decompose any (δ, τ)-lattice into a ﬁnite number of “big” separated subsequences.

Lemma 2.7. Let τ ∈ L and δ ∈ (0, mτ ). Let m be a positive integer. Any (δ, τ)-lattice {zj} on D can be partitioned into M subsequences such that, if aj and ak are diﬀerent m points in the same subsequence, then dτ (aj, ak) ≥ 2 δ.

Proof. Let K be the number given by Lemma 2.6. From the lattice {zj} extract a maximal m (2 δ)-subsequence, that is, we select one point ξ1 in our lattice, and then we continue m selecting points ξn of the lattice so that dτ (ξn, ξ) ≥ 2 δ for all previous selected point ξ.

37 We stop once the subsequence is maximal, that is, when all the remaining points x of the m lattice satisfy dτ (x, ξx) < 2 δ for some ξx in the subsequence. With the remaining points of the lattice we extract another maximal (2mδ)-subsequence, and we repeat the process until we get M = K + 1 maximal (2mδ)-subsequences. If no point of the lattice is left, we are done. On the other hand, if a point ζ in the lattice is left, this means that there are M = K + 1 distinct points xζ (at least one for each subsequence) in the lattice with m dτ (ζ, xζ ) < 2 δ, in contradiction with the choice of K from Lemma 2.6. The proof is complete.

Now we are ready for the main result of this Section, that characterizes the membership in the Schatten ideals of the Toeplitz operator acting on A2(ω). Theorem 2.8. Let ω ∈ W and 0 < p < ∞. The following conditions are equivalent: 2 (a) The Toeplitz operator Tµ is in Sp(A (ω)).

p (b) The function µbδ is in L (D, dλτ ) for δ ∈ (0, mτ ) suﬃciently small. p (c) The sequence {µbδ(zn)} is in ` for any δ > 0 small enough. Moreover, when p ≥ 1, the previous conditions are also equivalent to: p (d) The function Bωµ is in L (D, dλτ ). Proof. By Proposition 2.3 and the remark following it, the statements (b) and (c) are equivalent, and when p ≥ 1, they are also equivalent with condition (d). Hence, according to Proposition 2.5, the result is proved for p = 1 and we have the implication (a) implies (b) for p > 1. Moreover, the implication (c) implies (a) for 0 1, and that (a) implies (c) when 0 < p < 1.

p Let 1 < p < ∞, and assume that µbδ ∈ L (D, dλτ ) with δ ∈ (0, mτ ) small enough. It is not diﬃcult to see, using the equivalent discrete condition in (c) together with Theorem 2 2.1, that Tµ must be compact. For any orthonormal set {en} of A (ω), we have Z p X p X 2 hTµen, eniω = |en(z)| ω(z) dµ(z) . (2.5) n n D By Lemma A and Fubini’s theorem, Z Z Z 2 1 2 |en(z)| ω(z) dµ(z) . 2 |en(ζ)| ω(ζ) dA(ζ) dµ(z) D D τ(z) D(δτ(z))

Z 2 . |en(ζ)| ω(ζ) µbδ(ζ) dA(ζ). D 38 2 Since p > 1 and kenkAω = 1, we can apply H¨older’sinequality to get Z p Z 2 2 p |en(z)| ω(z) dµ(z) . |en(ζ)| ω(ζ) µbδ(ζ) dA(ζ). D D

2 −2 Putting this into (2.5) and taking into account that kKζ kA2(ω) ω(ζ) τ(ζ) , we see that

Z X p X 2 p hTµen, eniω . |en(ζ)| ω(ζ) µbδ(ζ) dA(ζ) n D n

Z 2 p ≤ kKζ k 2 ω(ζ) µδ(ζ) dA(ζ) Aω b D

Z p µbδ(ζ) dλτ (ζ). D

By [77, Theorem 1.27] this proves that T is in S with kT k kµ k p . µ p µ Sp . bδ L (D,dλτ )

2 Next, let 0 < p < 1, and suppose that Tµ ∈ Sp(A (ω)). We will prove that (c) holds. The method for this proof has his roots in previous work of S. Semmes [61] and D. Luecking p [39]. Let {zn} be a (δ, τ)-lattice on D. We want to show that {µbδ(zn)} is in ` . To this end, we ﬁx a large positive integer m ≥ 2 and apply Lemma 2.7 to partition the lattice {zn} into M subsequences such that any two distinct points aj and ak in the same subsequence m satisfy dτ (aj, ak) ≥ 2 δ. Let {an} be such a subsequence and consider the measure X ν = µχn, n where χn denotes the characteristic function of D(δτ(an)). Since m ≥ 2, the disks D(δτ(an)) are pairwise disjoints. Since Tµ is in Sp and 0 ≤ ν ≤ µ, then 0 ≤ Tν ≤ Tµ which implies that Tν is also in Sp. Moreover, kTνkSp ≤ kTµkSp . Fix an orthonormal basis 2 2 {en} for A (ω) and deﬁne an operator B on A (ω) by ! X X B λn en = λn fan , n n

where fan = Fan,N /τ(an) and Fan,N are the functions appearing in Lemma C with N taken big enough so that 3Np−4 > 2p. By [51, Proposition 2], the operator B is bounded. Since ∗ Tν ∈ Sp, the operator T = B TνB is also in Sp, with

2 kT kSp ≤ kBk · kTνkSp .

39 We split the operator T as T = D + E, where D is the diagonal operator on A2(ω) deﬁned by ∞ X 2 Df = hT en, eniω hf, eniω en, f ∈ A (ω), n=1 and E = T − D. By the triangle inequality,

kT kp ≥ kDkp − kEkp . (2.6) Sp Sp Sp

Since D is positive diagonal operator, we have X X kDkp = hT e , e ip = hT f , f ip Sp n n ω ν an an ω n n

Z p X 2 = |fan (z)| ω(z) dν(z) n D

Z 2 p X |Fa ,N (z)| ≥ n ω(z) dµ(z) . τ(a )2 n D(δτ(an)) n

Hence, by Lemma C, there is a positive constant C1 such that X kDkp ≥ C µ (a )p. (2.7) Sp 1 bδ n n On the other hand, since 0 < p < 1, by [77, Proposition 1.29] and Lemma 2.2 we have X X X kEkp ≤ hEe , e ip = hT f , f ip Sp n k ω ν an ak ω n k n,k:k6=n

p X Z ≤ |f (ξ)| |f (ξ)| ω(ξ) dν(ξ) an ak (2.8) n,k:k6=n D

p X X Z = |fan (ξ)| |fak (ξ)| ω(ξ) dµ(ξ) . n.k:k6=n j D(δτ(aj ))

m If n 6= k, then dτ (an, ak) ≥ 2 δ. Thus, for ξ ∈ D(δτ(aj)), is not diﬃcult to see that either m−2 m−2 dτ (ξ, an) ≥ 2 δ or dτ (ξ, ak) ≥ 2 δ.

40 m m Indeed, since n 6= k, then either dτ (an, aj) ≥ 2 δ or dτ (aj, ak) ≥ 2 δ. Suppose that m m−2 dτ (an, aj) ≥ 2 δ. If dτ (ξ, an) < 2 δ, then

m−2 |an − aj| ≤ |an − ξ| + |ξ − aj| < 2 δ min(τ(an), τ(ξ)) + δτ(aj) m−1 ≤ 2 δ min(τ(an), τ(aj)) + δτ(aj).

This directly gives a contradiction if min(τ(an), τ(aj)) = τ(aj). In case that min(τ(an), τ(aj)) = τ(an), using the Lipschitz condition (B) we get

m−1 |an − aj| < 2 δ min(τ(an), τ(aj)) + δτ(an) + c2δ|an − aj|

Since c2δ ≤ 1/4, and m ≥ 2, we see that this implies 4 d (a , a ) < (2m−1 + 1)δ ≤ 2mδ. τ n j 3

m−2 Thus, without loss of generality, we assume that dτ (ξ, an) ≥ 2 δ. For any n and k we write

X Z Ink(µ) = |fan (ξ)| |fak (ξ)| ω(ξ) dµ(ξ). j D(δτ(aj ))

With this notation and taking into account (2.8), we have X kEkp ≤ I (µ)p. (2.9) Sp nk n.k:k6=n By Lemma C, we have

−3N |Fan,N (ξ)| . dτ (ξ, an) .

m−2 Apply this inequality raised to the power 1/2, together with the fact that dτ (ξ, an) ≥ 2 δ to get

1/2 1/2 |Fan,N (ξ)| 1/2 − 3Nm |fan (ξ)| 2 (2.10) |fan (ξ)| = |Fan,N (ξ)| . 2 1/2 . τ(an) τ(an)

We also have 1/2 1/2 |fak (ξ)| ≤ |fak (ξ)| · kKξkA2(ω) (2.11)

Putting (2.10) and (2.11) into the deﬁnition of Ink(µ), and using the norm estimate

−1 −1/2 kKξkA2(ω) τ(ξ) ω(ξ) ,

41 we obtain

− 3Nm 2 2 X 1 Z I (µ) |f (ξ)|1/2 |f (ξ)|1/2 ω(ξ)1/2 dµ(ξ). nk . τ(a )1/2 τ(a )1/2 an ak n j j D(δτ(aj ))

By Lemma A, for ξ ∈ D(δτ(aj)), one has

Z 1/p 1/2 1/4 1 p/2 p/4 |fan (ξ)| ω(ξ) . 2 |fan (z)| ω(z) dA(z) τ(ξ) D(δτ(ξ))

−2/p 1/p . τ(aj) Sn(aj) , with Z p/2 p/4 Sn(x) = |fan (z)| ω(z) dA(z). D(3δτ(x)) In the same manner we also have

1/2 1/4 −2/p 1/p |fak (ξ)| ω(ξ) . τ(aj) Sk(aj) .

Therefore, there is a positive constant C2 such that

− 3Nm −4/p 2 2 X τ(aj) I (µ) ≤ C · S (a )1/p · S (a )1/p µ D(δτ(a )) nk 2 τ(a )1/2 τ(a )1/2 n j k j j n j j

− 3Nm 2 2 X = C · τ(a )3/2−4/p · S (a )1/p · S (a )1/p · µ (a ). 2 τ(a )1/2 j n j k j bδ j n j

Since 0 < p < 1 and 2−3Nm/2 ≤ 2−m, we get

−mp p p 2 X 3p −4 p I (µ) ≤ C · τ(a ) 2 · S (a ) · S (a ) · µ (a ) . nk 2 τ(a )p/2 j n j k j bδ j n j

Bearing in mind (2.9), this gives

p p −mp X 3p −4 p X Sn(aj) X kEk ≤ C · 2 τ(a ) 2 µ (a ) · S (a ) . Sp 2 j bδ j p/2 k j (2.12) τ(an) j n k

42 On the other hand, we have Z X X p/2 p/4 Sk(aj) = |fak (z)| ω(z) dA(z) k k D(3δτ(aj )) (2.13) Z X p/2 −p/2 p/4 = |Fak,N (z)| τ(ak) ω(z) dA(z). D(3δτ(aj )) k

Now, we claim that X τ(a )−p/2 |F (z)|p/2 τ(z)−p/2 ω(z)−p/4 . k ak,N . (2.14) k

In order to prove (2.14), note ﬁrst that using the estimate (1.8) in Lemma C, the estimate (1.2) and (iv) of Lemma B, we deduce that

X −p/2 p/2 τ(ak) |Fak,N (z)| {a ∈D(δ τ(z))} k 0 (2.15) −p/4 X −p/2 −p/2 −p/4 . ω(z) τ(ak) . τ(z) ω(z) . {ak∈D(δ0τ(z))} On the other hand, an application of the estimate (1.9) in Lemma C gives

X −p/2 p/2 τ(ak) |Fak,N (z)|

{ak∈/D(δ0τ(z))} 2 3Np p X τ(ak) −p/4 2 − 2 −2 . ω(z) τ(z) 3Np/2 |z − ak| {ak∈/D(δ0τ(z))} ∞ 2 3Np p X X τ(ak) −p/4 2 − 2 −2 = ω(z) τ(z) 3Np/2 , |z − ak| j=0 ak∈Rj (z) where j j+1 Rj(z) = ζ ∈ D : 2 δ0τ(z) < |ζ − z| ≤ 2 δ0τ(z) , j = 0, 1, 2 ...

Now observe that, using condition (B) in the deﬁnition of the class L, it is easy to see that, for j = 0, 1, 2,... ,

j j+1 D(δ0τ(ak)) ⊂ D(5δ02 τ(z)) if ak ∈ D(2 δ0τ(z)). This fact together with the ﬁnite multiplicity of the covering (see Lemma B) gives

X 2 j 2j 2 τ(ak) . m D(5δ02 τ(z)) . 2 τ(z) . ak∈Rj (z)

43 Therefore, as 3Np − 4 > 2p,

∞ X p X 3Npj X −p/2 p/2 −p/4 − 2 −2 − 2 2 τ(ak) |Fak,N (z)| . ω(z) τ(z) 2 τ(ak) {ak∈/D(δ0τ(z))} j=0 ak∈Rj (z) ∞ −p/4 −p/2 X (4−3Np) j . ω(z) τ(z) 2 2 j=0 −p/4 −p/2 . ω(z) τ(z) , which together with (2.15), proves the claim (2.14).

Putting (2.14) into (2.13) gives

X 2−p/2 Sk(aj) . τ(aj) . k

Similarly,

X Sn(aj) τ(a )2−p. τ(a )p/2 . j n n Putting these estimates into (2.12) we ﬁnally get X kEkp ≤ Cp · C · 2−mp µ (a )p Sp 2 3 bδ j j for some positive constant C3. Combining this with (2.6), (2.7) and choosing m large enough so that p −mp C2 · C3 · 2 ≤ C1/2, then we deduce that

X p C1 p p µδ(aj) ≤ kT k ≤ C4kTµk . b 2 Sp Sp j

Since this holds for each one of the M subsequences of {zn}, we obtain X µ (z )p ≤ C M kT kp (2.16) bδ n 4 µ Sp n for all locally ﬁnite positive Borel measures µ such that

X p µbδ(zn) < ∞. n Finally, an easy approximation argument then shows that (2.16) actually holds for all locally ﬁnite positive Borel measures µ. The proof is complete.

44 Chapter 3

Area operators

Let µ be a ﬁnite positive Borel measure on the unit disk D. The area operator Aµ is the sublinear operator deﬁned by

Z dµ(z) Aµ(f)(ζ) = |f(z)| 2 , ζ ∈ T := ∂D, Γ(ζ) 1 − |z|

where Γ(ζ) is a typical non-tangential approach region (Stolz region) with vertex ζ ∈ T. The study of the area operator acting on the classical Hardy spaces Hp was initiated p p by W. Cohn [14]. He proved that, for 0 < p < ∞, the area operator Aµ : H → L (T) is bounded if and only if µ is a classical Carleson measure (that is, a 1-Carleson measure according to our deﬁnitions in Chapter 1). The area operator is useful in harmonic analysis, and is closely related to, for example, non-tangential maximal functions, Poisson integrals, Littlewood-Paley operators, tent spaces, etc. The study of Aµ acting on Hardy spaces was p q pursued later in [24], where a full description of the boundedness of Aµ : H → L (T) for the case 0 < p < q < ∞ and 1 ≤ q < p < ∞ was obtained. In the setting of standard Bergman spaces, in [69] Z. Wu obtained a characterization of the boundedness p q of Aµ : Aα → L (T) for 1 ≤ p, q < ∞. In this chapter we are going to extend these results to our large Bergman spaces Ap(ω) for weights ω in the class W and characterize those p q positive Borel measures µ for which the area operator Aµ : A (ω) → L (T), 1 ≤ p, q < ∞ is bounded. The case q = 1 is also included in the other results of the chapter, but since it is easier and very instructive, it seems reasonable to look at it independently. For δ > 0 we use the notation 1 Z µ (z) = ω(ζ)−1/p dµ(z). bδ,p 2 τ(z) D(δτ(z))

Theorem 3.1. Let ω ∈ W and µ a ﬁnite positive Borel measure on D. Then

p 1 (i) Let 1 < p < ∞. Then Aµ : A (ω) → L (T) is bounded if and only if µbδ,p ∈ Lp/(p−1)(D).

45 p 1 (ii) Let 0 < p ≤ 1. Then Aµ : A (ω) → L (T) is bounded if and only if Z 1 −1/p sup 2/p ω(ζ) dµ(z) < ∞. z∈D τ(z) D(δ(τ(z))

Proof. By Fubini’s theorem, Z Z dµ(z) Z 1 kAµkL (T) = |f(z)| 2 dζ = |f(z)| dµ(z). T Γ(ζ) 1 − |z| D

p 1 p Thus Aµ : A (ω) → L (T) is bounded if and only if µ is a 1-Carleson measure for A (ω), and by Theorem 1 of [51] (see also Theorems D and F) this is equivalent to Z 1 −1/p sup 2/p ω(ζ) dµ(z) < ∞ if 0 < p ≤ 1, z∈D τ(z) D(δ(τ(z))

p/(p−1) and to µbδ,p ∈ L (D) if p > 1.

3.1 The case p ≤ q

For a positive function g deﬁned on the unit circle T, let 1 Z T g(z) = g(λ) |dλ|, 1 − |z| I(z) where I(0) = T, and for z 6= 0, I(z) denotes the arc in T with center z/|z| and length 2(1 − |z|). We also use the notation P g for the Poisson integral of g.

Lemma 3.2. For all z ∈ D, one has T g(z) ≤ C P g(z). Proof. Since |λ − z| ≤ C(1 − |z|) for λ ∈ I(z), one has Z 1 − |z|2 Z 1 − |z|2 P g(z) = g(λ) 2 |dλ| ≥ g(λ) 2 |dλ| ≥ C T g(z). T |λ − z| I(z) |λ − z|

p q Theorem 3.3. Let 1 0 suﬃciently small, the measure

p0 µbδ,p(z) dA(z) is a (p0/q0)-Carleson measure. Here we recall that 1 Z µ (z) = w(ζ)−1/pdµ(ζ). bδ,p 2 τ(z) D(δτ(z))

46 p q Proof. If 1 < q < ∞, by duality, the area operator Aµ : A (ω) → L (µ) is bounded if and only if there is a positive constant C such that Z A (f)(ζ) g(ζ) |dζ| ≤ C kgk q0 kfk p µ L (T) A (ω) T 0 for each positive function g ∈ Lq (T), where q0 is the conjugate exponent of q. An application of Fubini’s theorem yields Z Z Z dµ(z) Aµ(f)(ζ) g(ζ)|dζ| = |f(z)| 2 g(ζ) |dζ| T T Γ(ζ) 1 − |z| Z Z dµ(z) = χΓ(ζ)(z)|f(z)| 2 g(ζ) |dζ| T D 1 − |z| Z 1 Z = 2 g(ζ) dζ |f(z)| dµ(z) D 1 − |z| I(z) Z = T g(z) |f(z)| dµ(z). D p q By Theorem F, we have that Aµ : A (ω) → L (T) is bounded if and only if for δ > 0 suﬃciently small, one has Z Z p0 1 −1/p p0 |T g(ζ)| ω(ζ) dµ(ζ) dA(z) ≤ Ckgk 0 . (3.1) 2 Lq (T) D τ(z) D(δτ(z)) Now, in order to prove the suﬃciency, we use Lemma 3.2, the fact that P g(ζ) ≤ CP g(z) for ζ ∈ D(δτ(z)), and Carleson-Duren’s theorem to obtain Z Z p0 1 −1/p 2 |T g(ζ)| ω(ζ) dµ(ζ) dA(z) D τ(z) D(δτ(z)) Z p0 p0 p0 ≤ C |P g(z)| µδ,p(z) dA(z) ≤ Ckgk 0 . b Lq (T) D

To prove the necessity, for a given arc I ⊂ T, take g = χI in (3.1) to obtain p0 Z 1 Z p0 −1/p q0 2 |T χI (ζ)| ω(ζ) dµ(ζ) dA(z) ≤ C|I| , S(I) τ(z) D(δτ(z))

and, since T χI (ζ) ≥ c > 0 for ζ ∈ D(δτ(z)) and z ∈ S(I), the result follows.

Theorem 3.4. Let 0 0 suﬃciently small,

q−1 q Z (1 − |a|) −1/p sup 2/p ω(z) dµ(z) < ∞. (3.2) a∈D τ(a) D(δτ(a))

47 p q Proof. Suppose ﬁrst that Aµ : A (ω) → L (T) is bounded. Fix a ∈ D, and consider the test function Fa,p(z) given in Lemma C. One has

2/p kAµ(Fa,p)kLq(T) ≤ CkFa,pkAp(ω) ≤ C τ(a) . On the other hand, there is a number δ > 0 (independent of a and ζ) with D(δτ(a)) ⊂ Γ(ζ) 1 for ζ ∈ 2 I(a) (if I is an arc, then λI denotes the arc with the same center as I and length −1/p λ|I|). Therefore, using the fact that |Fa,p(z)| ω(z) for z ∈ D(δτ(a)) (see Lemma C) one has Z Z Z q q q dµ(z) kAµ(Fa,p)kLq( ) ≥ (AµFa,p(ζ)) |dζ| = |Fa,p(z)| |dζ| T 1 1 1 − |z| 2 I(a) 2 I(a) Γ(ζ) Z Z dµ(z) q ≥ |Fa,p(z)| |dζ| 1 1 − |z| 2 I(a) D(δτ(a)) Z dµ(z) q ≥ C|I(a)| ω(z)−1/p D(δτ(a)) 1 − |z| Z q ≥ C(1 − |a|)1−q ω(z)−1/p dµ(z) . D(δτ(a)) Thus 1 1− q Z (1 − |a|) −1/p sup 2/p ω(z) dµ(z) < ∞. a∈D τ(a) D(δτ(a))

p Conversely, suppose that (3.2) holds. We must show that the area operator Aµ : A (ω) → q L (T) is bounded. Let δ > 0 be suﬃciently small, and let {zj} be a (δ, τ)-lattice on D. We use the notation Dj = D(δτ(zj)) and Dej = De(δτ(zj)). Using Lemma A we obtain Z dµ(z) Aµf(ζ) = |f(z)| Γ(ζ) 1 − |z| Z Z 1/p 1 p dµ(z) ≤ C 2 |f(ξ)| ω(ξ) dA(ξ) Γ(ζ) ω(z) τ(z) D(δτ(z)) 1 − |z| Z −1/p Z 1/p X ω(z) p dµ(z) ≤ C 2/p |f(ξ)| ω(ξ) dA(ξ) Dj τ(z) D(δτ(z)) 1 − |z| j:Dj ∩Γ(ζ)6=∅ 1/p Z ! −1 Z X p (1 − |zj|) −1/p ≤ C |f(ξ)| ω(ξ) dA(ξ) 2/p ω(z) dµ(z) Dej τ(zj) Dj j:Dj ∩Γ(ζ)6=∅

Note that in the last inequality we used that τ(z) τ(zj) for z ∈ Dj. This, together with the assumption (3.2) yields

Z !1/p X p −1/q Aµf(ζ) ≤ C |f(ξ)| ω(ξ) dA(ξ) (1 − |zj|) . Dej j:Dj ∩Γ(ζ)6=∅

48 Since 0 < p ≤ 1, and (1 − |zj|) (1 − |ξ|) for ξ ∈ Dej, we get Z p X p −p/q Aµf(ζ) ≤ C |f(ξ)| (1 − |ξ|) ω(ξ) dA(ξ) Dej j:Dj ∩Γ(ζ)6=∅ Z ≤ C |f(ξ)|p (1 − |ξ|)−p/q ω(ξ) dA(ξ), Γ(e ζ)

where Γ(e ζ) is some Stolz angle with vertex at ζ with bigger aperture than Γ(ζ). Thus, by H¨older’sinequality and Fubini’s theorem, Z q pq/p kAµfkLq = Aµf(ζ) |dζ| T Z Z q/p ≤ C |f(ξ)|p (1 − |ξ|)−p/q ω(ξ) da(ξ) |dζ| T Γ(e ζ) Z Z q−p p dA(ξ) ≤ CkfkAp(ω) |f(ξ)| ω(ξ) |dζ| T Γ(e ζ) 1 − |ξ| Z Z dA(ξ) = Ckfkq−p |f(ξ)|p ω(ξ) χ (ξ) |dζ| Ap(ω) Γ(e ζ) D T 1 − |ξ| q ≤ CkfkAp(ω).

Note that the last inequality is due to the fact that R χ (ξ) |dζ| (1 − |ξ|). Thus the T Γ(e ζ) proof is complete.

3.2 The case q < p

p Theorem 3.5. Let 1 ≤ q 0 suﬃciently small, the function

Z p0 µbδ,p(z) Fµ(ζ) = dA(z) (3.3) Γ(ζ) 1 − |z|

q(p−1) belongs to L p−q (T).

1 Proof. Observe that, by Fubini’s theorem, the condition Fµ ∈ L (T) is equivalent to the p/(p−1) function µbδ,p being in L (D). Thus, the case q = 1 is just part (i) of Theorem 3.1. Now, suppose that 1 < q < ∞. In the same way as in the proof of Theorem 3.3, we see p q q0 that Aµ : A (ω) → L (T) is bounded if and only if for any positive function g ∈ L (T) we have Z Z p0 1 −1/p p0 |T g(ζ)| ω(ζ) dµ(ζ) dA(z) ≤ Ckgk 0 . (3.4) 2 Lq (T) D τ(z) D(δτ(z))

49 q0 q(p−1) Observe that q0−p0 = p−q . As |T g(ζ)| . |P g(ζ)| and |P g(ζ)| |P g(z)| for ζ ∈ D(δτ(z)), if condition (3.3) holds, since now p0 < q0, an application of Theorem C gives

Z Z p0 1 −1/p 2 |T g(ζ)| ω(ζ) dµ(ζ) dA(z) D τ(z) D(δτ(z))

Z p0 p0 p0 ≤ C |P g(z)| µδ,p(z) dA(z) ≤ Ckgk 0 . b Lq (T) D For the converse, we use that ∗ P h(z) . T (P h) )(z) (3.5) ∗ where u (ζ) = supz∈Γ(ζ) |u(z)| is the non-tangential maximal function of u. To prove (3.5), simply use the obvious inequality

P h(z) ≤ (P h)∗(ζ), ζ ∈ I(z)

and then integrate respect to ζ on I(z). Hence, applying (3.4) with g = (P h)∗ we obtain

Z Z p0 1 −1/p p0 |P h(ζ)| ω(ζ) dµ(ζ) dA(z) ≤ Ckhk 0 . 2 Lq (T) D τ(z) D(δτ(z)) As |P h(ζ)| |P h(z)| for ζ ∈ D(δτ(z)), this gives Z p0 p0 p0 |P h(z)| µδ,p(z) dA(z) ≤ Ckhk 0 . b Lq (T) D By Theorem C we see that (3.3) is satisﬁed.

Finally, in the case that 0 < q < 1 we have a suﬃcient condition. We didn’t know if this condition is also necessary, but when q = 1 it coincides with the one that characterizes the boundedness of the area operator. Theorem 3.6. Let q < p < ∞ with 0 < q ≤ 1. Suppose that the function

1−q q Z µ (1 − |z|) −1/p Fp,q(z) = 2/q ω(ζ) dµ(ζ) τ(z) D(δτ(z))

pq p q belongs to L p−q (D). Then the area operator Aµ : A (ω) → L (T) is bounded. Proof. As can be seen from the proof of Theorem 3.4, we have

!1/p Z 1 (q−p) X p 2 − q −2 pq µ Aµf(ζ) . |f(z)| ω(z) dA(z) (1 − |zj| ) τ(zj) Fp,q(zj). Dej Dj ∩Γ(ζ)6=∅

50 Since 0 < q ≤ 1, this gives !q/p Z (q−p) q X p 2 −1 −2 p µ q Aµf(ζ) . |f(z)| ω(z) dA(z) (1 − |zj| ) τ(zj) Fp,q(zj) . Dej Dj ∩Γ(ζ)6=∅ At this point we use H¨older’sinequality in order to obtain q q/p (p−q)/p Aµf(ζ) . (I) · (II) , (3.6) with Z X p −1 (I) := |f(z)| ω(z) dA(z) (1 − |zj|) Dej Dj ∩Γ(ζ)6=∅ X Z dA(z) |f(z)|p ω(z) (3.7) Dej 1 − |z| Dj ∩Γ(ζ)6=∅ Z p dA(z) . |f(z)| ω(z) , Γ(e ζ) 1 − |z| where Γ(e ζ) is a Stolz region with bigger aperture, and qp X 2 −1 2 µ p−q (II) := (1 − |zj| ) τ(zj) Fp,q(zj) .

Dj ∩Γ(ζ)6=∅ µ µ Now, it is clear that Fp,q(zj) Fp,q(z) if z ∈ Dj. Thus

qp Z qp 2 µ p−q µ p−q τ(zj) Fp,q(zj) Fp,q(z) dA(z). Dj This gives

Z qp dA(z) Z qp dA(z) X µ p−q µ p−q (II) Fp,q(z) . Fp,q(z) . (3.8) Dj 1 − |z| Γ(e ζ) 1 − |z| Dj ∩Γ(ζ)6=∅ Finally, bearing in mind (3.6), (3.7) and (3.8), we get Z kA fkq = A f(ζ)q|dζ| µ Lq(T) µ T q/p p−q Z Z dA(z) Z qp dA(z) p p µ p−q . |f(z)| ω(z) Fp,q(z) |dζ| T Γ(e ζ) 1 − |z| Γ(e ζ) 1 − |z| q/p p−q Z Z dA(z) Z Z qp dA(z) p p µ p−q ≤ |f(z)| ω(z) |dζ| Fp,q(z) |dζ| T Γ(e ζ) 1 − |z| T Γ(e ζ) 1 − |z| q/p p−q Z Z qp p p µ p−q |f(z)| ω(z) dA(z) Fp,q(z) dA(z) D D This ﬁnishes the proof.

51 52 Chapter 4

The exponential weight

For the case of the exponential weight A ω(z) = exp − , A > 0 1 − |z|2

a further study of the corresponding weighted Bergman space can be done, as it has been recently proved in [16] that the corresponding reproducing kernel Kz satisﬁes the estimate Z 1/2 −1/2 |Kz(ξ)| ω(ξ) dA(ξ) . ω(z) . (4.1) D This estimate allows to study the non-Hilbert space case. It turns out that when studying properties or operators such as the Bergman projection, Toeplitz operators or Hankel operators, where the reproducing kernels are involved, the most convenient settings are the spaces Ap(ωp/2). We consider the class E to consist of those weights in W satisfying condition (4.1). In the last section of this chapter, we will prove that the family of exponential type weights ωσ given by (1.1) satisfy condition (4.1) for 0 < σ < ∞.

4.1 Integral estimates for reproducing kernels

For a given weight v we introduce the growth space L∞(v) that consists of those measurable functions f on D such that

kfkL∞(v) := ess sup |f(z)|v(z) < ∞, z∈D

and let A∞(v) := L∞(v) ∩ H(D).

Lemma 4.1. Let ω ∈ E. For each z ∈ D, we have

−1/2 −2 kKzkA∞(ω1/2) . ω(z) τ(z) .

53 Proof. By Lemma A and condition (4.1), we have

1/2 Z 1/2 1/2 ω(ξ) 1/2 ω(ξ) |Kz(ξ)| = ω(ξ) |Kξ(z)| . 1/2 2 |Kξ(s)| ω(s) dA(s) ω(z) τ(z) D(δτ(z)) −1/2 −2 . ω(z) τ(z) .

This Lemma together with condition (4.1) allows us to obtain the following estimate for the norm of the reproducing kernel in Ap(ωp/2).

Lemma 4.2. Let 1 ≤ p < ∞, ω ∈ E and z ∈ D. Then

−1/2 −2(p−1)/p kKzkAp(ωp/2) ω(z) τ(z) .

Proof. By (4.1) and Lemma 4.1, we have Z Z p p p/2 1/2 1/2p−1 kKzkAp(ωp/2) = |Kz(ξ)| ω(ξ) dA(ξ) = |Kz(ξ)| ω(ξ) |Kz(ξ)| ω(ξ) dA(ξ) D D

Z p−1 1/2 p−1 −1/2 ≤ kKzkA∞(ω1/2) |Kz(ξ)| ω(ξ) dA(ξ) . kKzkA∞(ω1/2) ω(z) D

−p/2 −2(p−1) . ω(z) τ(z) . On the other hand, by Z p p p/2 kKzkAp(ωp/2) ≥ |Kz(ξ)| ω(ξ) dA(ξ) D(δτ(z))

Z p p 1/2 ≥ CkKzkA2(ω) kKξkA2(ω) ω(ξ) dA(ξ) D(δτ(z))

≥ C ω(z)−p/2τ(z)−2(p−1).

4.2 Bounded projections

The boundedness of Bergman projection is a fact of fundamental importance. In the case of the unit disc, the boundedness of Bergman projections is studied in [27], [77] and it

54 immediately gives the duality between the Bergman spaces. The orthogonal Bergman 2 2 projection Pω from L (D, ωdA) to A (ω) is given by Z Pωf(z) = f(ξ)Kz(ξ) ω(ξ) dA(ξ). D The natural Bergman projection is not necessarily bounded on Lp(D, ωdA) unless p = 2, (see [18] and [71] for more details). However, we are going to see next that Pω is bounded p p/2 p p/2 + on L (ω ) := L (D, ω dA). We ﬁrst prove the boundedness of the operator Pω deﬁned as Z + Pω f(z) = f(ξ) |Kz(ξ)| ω(ξ) dA(ξ). D We mention here that, for the case of the exponential weight with σ = 1, the results of this section has been obtained recently in [16].

+ p p/2 Theorem 4.3. Let 1 ≤ p < ∞ and ω ∈ E. The operator Pω is bounded on L (ω ) and on L∞(ω1/2).

Proof. We ﬁrst consider the easiest case p = 1. By Fubini’s theorem and condition (4.1) we obtain Z + + 1/2 kPω fkA1(ω1/2) = |Pω f(z)| ω(z) dA(z) ZD Z 1/2 ≤ |f(z)||Kz(ξ)| ω(ξ) ω(z) dA(ξ) dA(z) D D Z Z 1/2 = |f(ξ)| ω(ξ) |Kξ(z)| ω(z) dA(z) dA(ξ) D D Z 1/2 . |f(ξ)| ω(ξ) dA(ξ) = kfkL1(ω1/2). D Next, we consider the case 1 < p < ∞. Let p0 denote the conjugate exponent of p. By H¨older’sinequality and (4.1) we get

p−1 Z p+1 Z + p p 2 1/2 |Pω f(z)| ≤ |f(ξ)| |Kz(ξ)| ω(ξ) dA(ξ) |Kz(ξ)|ω(ξ) dA(ξ) D D

Z p p+1 − (p−1) . |f(ξ)| |Kz(ξ)|ω(ξ) 2 dA(ξ) ω(z) 2 . D

55 This together with Fubini’s theorem and another application of (4.1) gives Z + p + p p/2 kPω fkAp(ωp/2) = |Pω f(z)| ω(z) dA(z) D

Z Z p p+1 1/2 . |f(ξ)| |Kz(ξ)| ω(ξ) 2 dA(ξ) ω(z) dA(z) D D

Z Z p p+1 1/2 = |f(ξ)| ω(ξ) 2 |Kξ(z)| ω(z) dA(z) dA(ξ) D D

p . kfkLp(ωp/2).

Finally, if f ∈ L∞(ω1/2), by condition (4.1) we get Z 1/2 + 1/2 ω(z) |Pω (f)(z)| ≤ ω(z) |f(ξ)| |Kz(ξ)| ω(ξ)dA(ξ) D Z 1/2 1/2 ≤ kfkL∞(ω1/2) ω(z) |Kz(ξ)| ω(ξ) dA(ξ) D . kfkL∞(ω1/2).

+ ∞ 1/2 This shows that Pω is bounded on L (ω ). The proof is complete.

p p/2 Theorem 4.4. Let 1 ≤ p < ∞ and ω ∈ E. The Bergman projection Pω : L (ω ) −→ p p/2 ∞ 1/2 ∞ 1/2 A (ω ) is bounded. Moreover, Pω : L (ω ) → A (ω ) is also bounded.

Proof. In view of Theorem 4.3, it remains to see that Pωf deﬁnes an analytic function + on D. This follows easily by density, the boundedness of Pω and the completeness of Ap(ωp/2).

1 1/2 Corollary 4.5. Let ω ∈ E. The reproducing formula f = Pωf holds for each f ∈ A (ω ). Proof. This is an immediate consequence of the boundedness of the Bergman projection and the density of the polynomials.

56 4.3 Complex interpolation and duality

An elementary introduction to the basic theory of complex interpolation, including the complex interpolation of Lp spaces can be found in Chapter 2 of the book [77]. We assume in this section that the reader is familiar with that theory. First of all, we recall the following well-known interpolation theorem of Stein and Weiss [63].

Theorem 4.6. Suppose that ω, ω0 and ω1 are weight functions on D. If 1 ≤ p0 ≤ p1 ≤ ∞ and 0 ≤ θ ≤ 1, then

p0 p1 p L (D, ω0dA),L (D, ω1dA) θ = L (D, ωdA) with equal norms, where

1−θ θ 1 1 − θ θ 1/p p0 p1 = + , ω = ω0 ω1 . p p0 p1

With this and the result on bounded projections we can obtain the following result on complex interpolation of large weighted Bergman spaces.

Theorem 4.7. Let ω be a weight in the class E. If 1 ≤ p0 ≤ p1 ≤ ∞ and 0 ≤ θ ≤ 1, then

[Ap0 (ωp0/2),Ap1 (ωp1/2)] = Ap(ωp/2),

where 1 1 − θ θ = + . p p0 p1

Proof. The inclusion [Ap0 (ωp0/2),Ap1 (ωp1/2)] ⊂ Ap(ωp/2) is a consequence of the deﬁnition of complex interpolation, the fact that each Apk (ωpk/2) is a closed subspace of Lpk (ωpk/2) p0 p0/2 p1 p1/2 p p/2 and L (ω ),L (ω ) θ = L (ω ). This last assertion follows from Theorem 4.6. On the other hand, if f ∈ Ap(ωp/2) ⊂ Lp(ωp/2), it follows from Theorem 4.6 that

p0 p0/2 p1 p1/2 p p/2 L (ω ),L (ω ) θ = L (ω ).

Thus, there exists a function Fζ (z)(z ∈ D and 0 ≤ Re ζ ≤ 1) and a positive constant C such that:

(a) Fθ(z) = f(z) for all z ∈ D.

(b) kFζ kLp0 (ωp0/2) ≤ C for all Re ζ = 0.

Deﬁne a function Gζ by Gζ (z) = PωFζ (z). Due to the reproducing formula in Corollary 4.5 and Theorem 4.4 we have:

57 (a) Gθ(z) = f(z) for all z ∈ D.

(b) kGζ kLp0 (ωp0/2) ≤ C for all Re ζ = 0.

p0 p0/2 p1 p1/2 Since each function Gζ is analytic on D, we conclude that f belongs to [A (ω ),A (ω )]. This completes the proof of the theorem.

As in the case of the standard Bergman spaces, one can use the result just proved on p p/2 the boundedness of the Bergman projection Pω in L (ω ) to identify the dual spaces of Ap(ωp/2). As usual, if X is a Banach space, we denote its dual by X∗. Next two results (Theorems 4.8 and 4.9) on the duality of Bergman spaces with exponential type weights appears also on [16].

Theorem 4.8. Let ω ∈ E and 1 < p < ∞. The dual space of Ap(ωp/2) can be identiﬁed (with equivalent norms) with Ap0 (ωp0/2) under the integral pairing Z hf, giω = f(z) g(z) ω(z) dA(z). D Here p0 denotes the conjugate exponent of p, that is, p0 = p/(p − 1).

Proof. Let 1 < p < ∞ and let p0 = p/(p − 1) be its dual exponent. Given a function p0 p0/2 p p/2 g ∈ A (ω ), H¨older’sinequality implies that the linear functional ψg : A (ω ) −→ C deﬁned by Z p p/2 ψg(f) := f(ξ) g(ξ) ω(ξ) dA(ξ), f ∈ A (ω ), D

is bounded with kψgk ≤ kgkAp0 (ωp0/2). p p/2 ∗ Conversely, let T ∈ A (ω ) . By Hahn-Banach we can extend T to an element Te ∈ p p/2 ∗ L (D, ω dA) such that kTek = kT k. By the Riesz representation Theorem there exists p0 p/2 H ∈ L (D, ω dA) with kHkLp0 (ωp/2) = kTek = kT k such that Z Te(f) = f(ξ) H(ξ) ω(ξ)p/2 dA(ξ), D

p p/2 p −1 p0 p0/2 for every f ∈ A (ω ). Consider the function h(ξ) = H(ξ) ω(ξ) 2 . Then h ∈ L (ω ) with

khkLp0 (ωp0/2) = kHkLp0 (ωp/2) = kT k,

and Z T (f) = Te(f) = f(ξ) h(ξ) ω(ξ) dA(ξ), f ∈ Ap(ωp/2). D

58 p0 p0/2 p0 p0/2 Let g = Pωh. By Theorem 4.4, the Bergman projection Pω : L (ω ) −→ A (ω ) is bounded. Thus g ∈ Ap0 (ωp0/2) with

kgkAp0 (ωp0/2) = kPωhkAp0 (ωp0/2) . khkLp0 (ωp0/2) = kT k.

From Fubini’s theorem it is easy to see that Pω is self-adjoint. Indeed, Z hPωf, giω = Pωf(ξ) g(ξ) ω(ξ) dA(ξ) D

Z Z = f(s) Kξ(s) ω(s) dA(s) g(ξ) ω(ξ) dA(ξ) D D

Z Z = g(ξ) Ks(ξ) ω(ξ) dA(ξ) f(s) ω(s) dA(s) D D

Z = Pωg(s) f(s) ω(s) dA(s) = hf, Pωgiω. D The interchange of the order of integration is well justiﬁed, because of the boundedness of + the operator Pω (see Theorem 4.3) given by Z + Pω f(z) = f(ζ) |Kz(ζ)| ω(ζ) dA(ζ). D

p p/2 Therefore, since f = Pωf for every f ∈ A (ω ) according to Corollary 4.5, we get Z T (f) = f(ξ) h(ξ) ω(ξ) dA(ξ) D = hf, Pωhiω = hf, giω = ψg(f).

p0 p0/2 Finally, the function g is unique. Indeed, if there is another function ge ∈ A (ω ) with T (f) = ψ (f) = ψ (f) for every f ∈ Ap(ωp/2), then by taking f = K for each a ∈ (that g ge a D belongs to Ap(ωp/2) due to Lemma 4.2), and using the reproducing formula, we obtain

g(a) = ψ (K ) = ψ (K ) = g(a), a ∈ . g a ge a e D p0 p0/2 Thus, any bounded linear functional T is of the form T = ψg for some unique g ∈ A (ω ) and, furthermore

kT k kgkAp0 (ωp0/2). The proof is complete.

59 Theorem 4.9. Let ω ∈ E. The dual space of A1(ω1/2) can be identiﬁed (with equivalent ∞ 1/2 norms) with A (ω ) under the integral pairing hf, giω.

∞ 1/2 1 1/2 Proof. Let g ∈ A (ω ). The linear functional ψg : A (ω ) −→ C deﬁned by ψg(f) := 1 1/2 hf, giω is bounded with kψgk ≤ kgkL∞(ω1/2), since for every f ∈ A (ω )

|ψg(f)| ≤ kgkL∞(ω1/2)kfkA1(ω1/2).

Conversely, let T ∈ A1(ω1/2)∗ . Consider the space X that consists of the functions of the form h = fω1/2 with f ∈ A1(ω1/2). Clearly X is a subspace of L1(D, dA) and F (h) := T (hω−1/2) = T (f) deﬁnes a bounded linear functional on X with kF k = kT k. By ∗ Hahn- Banach, F has an extension Fe ∈ (L1(D, dA)) with kFek = kF k. Hence, there is a ∞ function G ∈ L (D, dA) with kGkL∞(D,dA) = kF k such that Z F (h) = Fe(h) = h(ξ) G(ξ) dA(ξ), h ∈ X, D or Z T (f) = f(ξ)G(ξ) ω(ξ)1/2 dA(ξ), f ∈ A1(ω1/2). D Consider the function H(z) = ω(z)−1/2G(z). Then H ∈ L∞(ω1/2) with

kHkL∞(ω1/2) = kGkL∞(D,dA) = kF k = kT k.

∞ 1/2 By Theorem 4.4, the function g = PωH is in A (ω ) with

kgkA∞(ω1/2) . kHkL∞(ω1/2) = kT k.

Also, for f ∈ A1(ω1/2), by the reproducing formula, we have Z T (f) = f(ξ)H(ξ) ω(ξ) dA(ξ) = hPωf, Hiω = hf, PωHiω = ψg(f). D Finally, as in the proof of Theorem 4.8 the function g is unique.

Corollary 4.10. Let ω ∈ E. The set E of ﬁnite linear combinations of reproducing kernels is dense in Ap(ωp/2), 1 ≤ p < ∞.

Proof. Since E is a linear subspace of Ap(ωp/2), by standard functional analysis and the duality results in Theorems 4.8 and 4.9, it is enough to prove that g ≡ 0 if g ∈ Ap0 (ωp0/2) 0 satisﬁes hf, giω = 0 for each f in E (with p being the conjugate exponent of p, and ∞ 1/2 g ∈ A (ω ) if p = 1). But, taking f = Kz for each z ∈ D and using the reproducing formula, we get g(z) = Pωg(z) = hg, Kziω = 0, for each z ∈ D. This ﬁnishes the proof.

60 Our next goal is to identify the predual of A1(ω1/2). For a given weight v, we introduce the ∞ space A0(v) consisting of those functions f ∈ A (v) with lim|z|→1− v(z)|f(z)| = 0. Clearly, ∞ A0(v) is a closed subspace of A (v).

1/2 Theorem 4.11. Let ω ∈ E. Under the integral pairing hf, giω, the dual space of A0(ω ) can be identiﬁed (with equivalent norms) with A1(ω1/2).

1 1/2 Proof. If g ∈ A (ω ), clearly Λg(f) = hf, giω deﬁnes a bounded linear functional in 1/2 1/2 ∗ A0(ω ) with kΛgk ≤ kgkA1(ω1/2). Conversely, assume that Λ ∈ A0(ω ) . Consider the 1/2 1/2 space X that consists of functions of the form h = fω with f ∈ A0(ω ). Clearly X is a subspace of C0(D) (the space of all continuous functions vanishing at the boundary) and T (h) = Λ(ω−1/2h) = Λ(f) deﬁnes a bounded linear functional on X with kT k = kΛk. ∗ By Hahn-Banach, T has an extension Te ∈ C0(D) with kTek = kT k. Hence, by Riesz representation theorem, there is a measure µ ∈ M(D) (the Banach space of all complex

Borel measures µ equipped with the variation norm kµkM ) with kµkM = kT k such that Z T (h) = Te(h) = h(ζ) dµ(ζ), h ∈ X, D or Z 1/2 1/2 Λ(f) = f(ζ) ω(ζ) dµ(ζ), f ∈ A0(ω ). D Consider the function g deﬁned on the unit disk by Z 1/2 g(z) = Kz(ζ) ω(ζ) dµ(ζ), z ∈ D. D Clearly g is analytic on D and, by Fubini’s theorem and condition (4.1), we have Z Z 1/2 1/2 kgkA1(ω1/2) ≤ |Kz(ζ)| ω(ζ) d|µ|(ζ) ω(z) dA(z) D D

Z Z 1/2 1/2 = |Kζ (z)| ω(z) dA(z) ω(ζ) d|µ|(ζ) D D

. |µ|(D) = kµkM = kΛk,

1 1/2 1/2 2 proving that g belongs to A (ω ). Now, since A0(ω ) ⊂ A (ω), the reproducing formula

61 1/2 f(ζ) = hf, Kζ iω holds for all f ∈ A0(ω ). This and Fubini’s theorem yields Z Z 1/2 Λg(f) = hf, giω = f(z) Kz(ζ) ω(ζ) dµ(ζ) ω(z) dA(z) D D

Z Z 1/2 = f(z) Kζ (z) ω(z) dA(z) ω(ζ) dµ(ζ) D D

Z = f(ζ) ω(ζ)1/2 dµ(ζ) = Λ(f). D By the reproducing formula, the function g is uniquely determined by the identity g(z) = Λ(Kz). This completes the proof.

For the case of normal weights, the analogues of Theorems 4.9 and 4.11 were obtained by Shields and Williams in [62]. They also asked what happens with the exponential weights, problem that is solved in the present work.

4.4 Atomic decomposition

With the help of the duality results and the estimates for the p-norm of the reproducing kernels Kz, we can obtain the atomic decomposition of Bergman spaces with exponential weights in the case that p ≥ 1. We use the notation kp,z for the normalized reproducing kernels in Ap(ωp/2), that is Kz kp,z = . kKzkAp(ωp/2)

Theorem 4.12. Let ω ∈ E and 1 ≤ p < ∞. There exists a lattice {zn} ⊂ D such that:

p (i) For any λ = {λn} ∈ ` , the function X f(z) = λn kp,zn (z) n

p p/2 is in A (ω ) with kfkAp(ωp/2) ≤ Ckλk`p .

p p/2 p (ii) For every f ∈ A (ω ) exists λ = {λn} ∈ ` such that X f(z) = λn kp,zn (z) n

and kλk`p ≤ C kfkAp(ωp/2).

62 Proof. (i) As done in Proposition 1.5, the function f deﬁnes an analytic function on D. Set ∞ X 2 1/2 M(z) := τ(zk) ω(zk) |Kzk (z)|. k=0 By H¨older’sinequality we have

∞ X 2 1/2 M(z) : = τ(zk) ω(zk) |Kzk (z)| k=0 ∞ Z X 1/2 . |Kz(ξ)| ω(ξ) dA(ξ) k=0 D(δτ(zk)) Z 1/2 −1/2 . |Kz(ξ)| ω(ξ) dA(ξ) . ω(z) . D Therefore, applying condition (4.1) again, we obtain Z ∞ ! p X p 1/2 1/2 kF kAp(ωp/2) . |λk| ω(zk) |Kzk (z)| ω(z) dA(z) D k=0 ∞ Z X p 1/2 1/2 . |λk| ω(zk) |Kzk (z)|ω(z) dA(z) k=0 D ∞ X p . |λk| . k=0

In order to prove (ii), we deﬁne a linear operator S : `p −→ Ap(ωp/2) given by

∞ X S({λn}) := λn kp,zn . n=0 By (i), the operator S is bounded. By the duality results obtained in the previous Chapter, when 1 < p < ∞, the adjoint operator S∗ : Ap0 (ωp0/2) → `p0 , where p0 is the conjugate exponent of p, is deﬁned by

∗ X ∗ hSx, fiω = hx, S fi` = xn (S f)n. n

63 for every x ∈ `p and f ∈ Ap0 (ωp0/2). From here, the proof follows the same lines as in Theorem 1.7

When p = 1, then S∗ : A∞(ω1/2) −→ `∞ is given by ∗ f(zn) {(S f)n} = . kKzn kA1(ω1/2) n Hence we must show that

1/2 ∗ 1/2 sup ω(z) |f(z)| = kfkA∞(ω1/2) . kS fk`∞ sup ω(zn) |f(zn)|, z∈D n for f ∈ A∞(ω1/2). However this can be proved with the same method of the proof of Lemma 1.6. Indeed, let z ∈ D. Then there is a point zk with z ∈ D(ετ(zk)). By Lemma A, we have Z 1/2 C1 1/2 1/2 ω(z) |f(z)| ≤ 2 2 |f(ζ)| ω(ζ) − |f(zk)| ω(zk) dA(ζ) ε τ(z) D(ετ(z))

1/2 + C1 |f(zk)| ω(zk) . As done in the proof of Lemma 1.6, we have 1 Z |f(ζ)| ω(ζ)1/2 − |f(z )| ω(z )1/2 ≤ C ε |f(ξ)| ω(ξ)1/2 dA(ξ) k k 2 2 τ(zk) D(3δ0τ(zk))

≤ C3 ε kfkA∞(ω1/2). Thus, putting this in the previous estimate, we obtain

1/2 1/2 ω(z) |f(z)| ≤ C4 ε kfkA∞(ω1/2) + C1 sup ω(zn) |f(zn)|. n

Finally, taking the supremum on z and ε > 0 small enough so that C4ε ≤ 1/2, we have

1/2 kfkA∞(ω1/2) . sup ω(zn) |f(zn)|. n The proof is complete.

4.5 Toeplitz operators

In this section we are going to extend the results in Chapter 2 to the non-Hilbert space setting, when the weight ω is in the class E. Concretely, we describe the boundedness of

64 p p/2 q q/2 the Toeplitz operators Tµ : A (ω ) → A (ω ) for 1 ≤ p, q < ∞. Recall that the Toeplitz operator Tµ is deﬁned by Z Tµf(z) = f(ξ) Kz(ξ) ω(ξ) dµ(ξ). D

Also, recall that, for δ ∈ (0, mτ ), the averaging function of µ on D is given by µ(D(δτ(z)) µ (z) := , z ∈ . bδ τ(z)2 D

Theorem 4.13. Let ω ∈ E, 1 ≤ p ≤ q < ∞ and µ be a ﬁnite positive Borel measure p p/2 q q/2 on D satisfying (2.1). Then Tµ : A (ω ) → A (ω ) is bounded if and only if for each δ ∈ (0, mτ ) suﬃciently small

2( 1 − 1 ) E(µ) = sup τ(z) q p µbδ(z) < ∞. (4.2) z∈D Moreover,

kTµkAp(ωp/2)→Aq(ωq/2) E(µ).

−1/2 −2(p−1)/p Proof. Since we have the estimate kKzkAp(ωp/2) ω(z) τ(z) , if we assume that p p/2 q q/2 the Toeplitz operator Tµ : A (ω ) −→ A (ω ) is bounded, then we obtain (4.2) with the same argument as in the proof of Theorem 2.1.

Conversely, we suppose that (4.2) holds. We ﬁrst prove that Z 1/2 −2( 1 − 1 ) −1/2 |Kz(ξ)| ω(ξ) τ(ξ) p q dµ(ξ) . E(µ) ω(z) . (4.3) D Indeed, by Lemma A, we have Z 1/2 1 1/2 |Kz(ξ)| ω(ξ) . 2 |Kz(s)| ω(s) dA(s). τ(ξ) δ D( 2 τ(ξ)) Then, by Fubini’s theorem, the fact that τ(s) τ(ξ) for s ∈ D(δτ(ξ)), and condition (4.1), we get Z Z 1/2 −2( 1 − 1 ) 1/2 2( 1 − 1 ) |Kz(ξ)| ω(ξ) τ(ξ) p q dµ(ξ) . |Kz(s)| ω(s) τ(s) q p µbδ(s) dA(s) D D

Z 1/2 . E(µ) |Kz(s)| ω(s) dA(s) D

−1/2 . E(µ) ω(z) .

65 This establishes (4.3). Now we proceed to prove that Tµ is bounded. If q > 1, by H¨older’s inequality, we obtain Z q q |Tµf(z)| ≤ |f(ξ)| |Kz(ξ)| ω(ξ) dµ(ξ) D

Z Z q−1 q q+1 2( 1 − 1 )(q−1) 1/2 −2( 1 − 1 ) ≤ |f(ξ)| ω(ξ) 2 |Kz(ξ)| τ(ξ) p q dµ(ξ) |Kz(ξ)| ω(ξ) τ(ξ) p q dµ(ξ) . D D Using (4.3), we have Z q q−1 q q+1 2( 1 − 1 )(q−1) − (q−1) |Tµf(z)| . E(µ) |f(ξ)| ω(ξ) 2 |Kz(ξ)| τ(ξ) p q dµ(ξ) ω(z) 2 . D If q = 1, this holds directly. By Fubini’s theorem and condition (4.1), we obtain Z q q q/2 kTµfkAq(ωq/2) = |Tµf(z)| ω(z) dA(z) D

Z Z q−1 q (q+1) 2(q−1)( 1 − 1 ) 1/2 . E(µ) |f(ξ)| ω(ξ) 2 τ(ξ) p q |Kξ(z)| ω(z) dA(z) dµ(ξ) D D

Z q−1 q q/2 2(q−1)( 1 − 1 ) . E(µ) |f(ξ)| ω(ξ) τ(ξ) p q dµ(ξ). D Consider the measure ν given by

q/2 2(q−1)( 1 − 1 ) dν(ξ) := ω(ξ) τ(ξ) p q dµ(ξ).

p p/2 q Since (4.2) holds, then by Theorem D, the identity Iν : A (ω ) −→ L (D, dν) is bounded. 1/q Moreover, kIνk . E(µ) . Therefore, Z q q−1 q q q kTµfkAq(ωq/2) . E(µ) |f(z)| dν(z) . E(µ) · kfkAp(ωp/2). (4.4) D This ﬁnishes the proof.

p p/2 q q/2 In order to describe the boundedness of Tµ : A (ω ) → A (ω ) when 1 ≤ q < p < ∞, we need ﬁrst an auxiliary result.

pq Proposition 4.14. Let ω ∈ E and 1 < q < p < ∞. If µbδ ∈ L p−q (D, dA), then Z Z q 1 1/2 q q Jδ,q := |f(ξ)| ω(ξ) dµ(ξ) dA(z) . kµδk pq · kfk p p/2 , 2 b L p−q ( ) A (ω ) D τ(z) D(δτ(z)) D for any f ∈ Ap(ωp/2).

66 Proof. For z ∈ D and ξ ∈ D(δτ(z)), by Lemma A, Lemma B and (1.2), we obtain Z 1/p 1/2 1 p p/2 |f(ξ)| ω(ξ) . 2 |f(s)| ω(s) dA(s) τ(ξ) δ D( 3 τ(ξ)) Z 1/p 1 p p/2 . 2 |f(s)| ω(s) dA(s) . τ(z) D(δτ(z)) This gives,

1 Z 1 Z 1/p |f(ξ)| ω(ξ)1/2dµ(ξ) µ (z) |f(s)|p ω(s)p/2dA(s) . 2 . bδ 2 τ(z) D(δτ(z)) τ(z) D(δτ(z)) Therefore,

Z 1 Z q/p J |f(s)|p ω(s)p/2dA(s) µ (z)q dA(z). δ,q . 2 bδ D τ(z) D(δτ(z)) Applying H¨older’sinequality

Z Z q/p 1 p p/2 q Jδ,q . |f(s)| ω(s) dA(s)dA(z) kµδk pq . (4.5) 2 b L p−q ( ) D τ(z) D(δτ(z)) D On the other hand, by Fubini’s theorem and τ(z) τ(s), for s ∈ D(δτ(z)), we have Z Z 1 p p/2 p 2 |f(s)| ω(s) dA(s) dA(z) . kfkAp(ωp/2). D τ(z) D(δτ(z)) Combining this with (4.5), we get

q q J kfk · kµ k pq . δ,q . Ap(ωp/2) bδ L p−q (D) The proof is complete.

Theorem 4.15. Let ω ∈ E, 1 ≤ q < p < ∞ and µ be a ﬁnite positive Borel measure on D satisfying (2.1). The following conditions are equivalent:

p p/2 q q/2 (i) The Toeplitz operator Tµ : A (ω ) −→ A (ω ) is bounded.

pq (ii) For each suﬃciently small δ > 0, µbδ ∈ L p−q (D, dA). Moreover, we have

kTµkAp(ωp/2)→Aq(ωq/2) kµδk pq . b L p−q (D)

67 p Proof. (i) =⇒ (ii) For an arbitrary sequence λ = {λk} ∈ ` , we consider the function

∞ p−1 X 1/2 2( p ) Gt(z) = λk rk(t) ω(zk) τ(zk) Kzk (z) , 0 < t < 1, k=0 where rk(t) is a sequence of Rademacher functions see [41] or Appendix A of [21] and {zk} is the sequence given in Lemma B. Because of the norm estimate

−1/2 −2(p−1)/p kKzkAp(ωp/2) ω(z) τ(z) given in Lemma 4.2, by part (i) of Theorem 4.12 we obtain

∞ !1/p X p kGtkAp(ωp/2) . |λk| . k=0

p p/2 q q/2 Thus, the boundedness of Tµ : A (ω ) −→ A (ω ) gives

q q q kTµGtkAq(ωq/2) . kTµk · kλk`p . In other words, we have

∞ Z p−1 q X 1/2 2( p ) q/2 q q λ r (t) ω(z ) τ(z ) T K (z) ω(z) dA(z) kT k · kλk p . k k k k µ zk . µ ` D k=0 Integrating with respect to t from 0 to 1, applying Fubini’s theorem and invoking Khin- chine’s inequality (see [41]), we obtain

∞ !q/2 Z p−1 X 2 4( p ) 2 q/2 q q B := |λk| ω(zk) τ(zk) |TµKzk (z)| ω(z) dA(z) . kTµk · kλk`p . D k=0

Let χk denote the characteristic function of the set D(3δτ(zk)). Since the covering {D(3δτ(zk))} of D has ﬁnite multiplicity N, we have ∞ p−1 Z X q q/2 2q( p ) q q/2 |λk| ω(zk) τ(zk) |TµKzk (z)| ω(z) dA(z) k=0 D(3δτ(zk)) ∞ Z p−1 (4.6) X q q/2 2q( p ) q q/2 = |λk| ω(zk) τ(zk) |TµKzk (z)| χk(z) ω(z) dA(z) D k=0 1− q ≤ max{1,N 2 } B.

Now, using Lemma A, yields

∞ p−1 X q q/2 2q( p )+2 q q/2 q q |λk| ω(zk) τ(zk) |TµKzk (zk)| ω(zk) . kTµk · kλk`p . k=0

68 2 2 On the other hand, since for small δ > 0, we have |Kzk (z)| kKzk kA (ω)kKzkA (ω) for every z ∈ D(δτ(zk)), applying Lemma D and (1.2), we have Z 2 |TµKzk (zk)| ≥ |Kzk (z)| ω(z) dµ(z) D(δτ(zk)) Z 2 2 kKzk kA2(ω) kKzkA2(ω) ω(z) dµ(z) D(δτ(zk)) −1 ω(zk) µbδ(zk) 2 . τ(zk) That is, − q q ω(z ) 2 µ (z ) |T K (z )|q ω(z )q/2 k bδ k . µ zk k k & 2q τ(zk) Therefore, ∞ 1 1 X q 2q( q − p ) q q q |λk| τ(zk) µbδ(zk) . kTµk · kλk`p . k=0 p Then, using the duality between `p/q and ` p−q we conclude that

∞ p p−q X 2q( 1 − 1 ) q pq τ(zk) q p µbδ(zk) . kTµk p−q , k=0 that means ∞ X 2 pq pq τ(zk) µbδ(zk) p−q . kTµk p−q . k=0 This is the discrete version of our condition. To obtain the continuous version, simply note that µbδ(z) . µb4δ(zk), z ∈ D(δτ(zk)). Then, Z ∞ Z ∞ pq X pq X 2 pq µbδ(z) p−q dA(z) ≤ µbδ(z) p−q dA(z) . τ(zk) µb4δ(zk) p−q . D k=0 D(δτ(zk)) k=0 This ﬁnishes the proof of this implication.

(ii) =⇒ (i) First we begin with the easiest case q = 1. By Fubini’s theorem and

69 condition (4.1), we have Z 1/2 kTµfkA1(ω1/2) = |Tµf(z)| ω(z) dA(z) D

Z Z 1/2 ≤ |f(ξ)| |Kz(ξ)| ω(ξ) dµ(ξ) ω(z) dA(z) D D (4.7) Z Z 1/2 = |f(ξ)| |Kξ(z)| ω(z) dA(z) ω(ξ) dµ(ξ) D D

Z 1/2 . |f(ξ)| ω(ξ) dµ(ξ). D Now, by using Theorem F with the measure given by

dν(ξ) := ω(ξ)1/2 dµ(ξ), it gives the desired result.

Finally, we study the case 1 < q < ∞. Let {zj} be the sequence given in Lemma B. Applying Lemma B and Lemma A, we obtain

∞ X Z |Tµf(z)| ≤ |f(ξ)| |Kz(ξ)| ω(ξ) dµ(ξ) j=0 D(δτ(zj )

∞ X Z 1 Z |f(ξ)| ω(ξ)1/2 |K (s)| ω(s)1/2 dA(s) dµ(ξ) . τ(ξ)2 z j=0 D(δτ(zj )) D(δτ(ξ))

∞ X Z dµ(ξ) Z |f(ξ)| ω(ξ)1/2 |K (s)| ω(s)1/2 dA(s). . τ(ξ)2 z j=0 D(δτ(zj )) D(3δτ(zj ))

Applying H¨older’sinequality, we get

q |Tµf(z)| . M(z) × N(z), where

∞ q X Z dµ(ξ) Z M(z) := |f(ξ)| ω(ξ)1/2 |K (s)| ω(s)1/2 dA(s), τ(ξ)2 z j=0 D(δτ(zj )) D(3δτ(zj ))

70 and ∞ Z q−1 X 1/2 N(z) := |Kz(s)| ω(s) dA(s) . j=0 D(3δτ(zj )) Furthermore, by Lemma B and condition (4.1), we have

Z q−1 1/2 1−q N(z) . |Kz(s)| ω(s) dA(s) . ω(z) 2 . D Thus

q q/2 1/2 |Tµf(z)| ω(z) . M(z) ω(z) .

This gives ∞ q X Z dµ(ξ) kT fkq |f(ξ)| ω(ξ)1/2 K(j), µ Aq(ωq/2) . τ(ξ)2 j=0 D(δτ(zj )) where Z Z 1/2 1/2 K(j) := |Kz(s)| ω(s) dA(s) ω(z) dA(z), D D(3δτ(zj )) which by Fubini’s theorem and condition (4.1)

2 K(j) . τ(zj) . Combining this with using (1.2) and Proposition 4.14, it shows that

∞ q X 1 Z kT fkq τ(z )2 |f(ξ)| ω(ξ)1/2 dµ(ξ) µ Aq(ωq/2) . j τ(z )2 j=0 j D(δτ(zj ))

Z Z q 1 1/2 . 2 |f(ξ)| ω(ξ) dµ(ξ) dA(z) D τ(z) D(4δτ(z))

q q kµ k pq · kfk . . b4δ Ap(ωp/2) L p−q (D) This proves the desired result.

Next we characterize compact Toeplitz operators on weighted Bergman spaces Ap(ωp/2) for weights ω in the class E. We need ﬁrst a lemma.

p p/2 Lemma 4.16. Let 1 < p < ∞, and let kp,z the normalized reproducing kernels in A (ω ). − Then kp,z → 0 weakly as |z| → 1 .

71 Proof. By duality and the reproducing kernel properties, we must show that |g(z)|/kKzkAp(ωp/2) goes to zero as |z| → 1− whenever g is in Ap0 (ωp0/2), where p0 denotes the conjugate exponent of p, but this follows easily by the density of the polynomials and Lemma A.

Theorem 4.17. Let ω ∈ E, 1 < p ≤ q < ∞ and µ be a ﬁnite positive Borel measure on D p p/2 q q/2 satisfying (2.1). Then the Toeplitz operator Tµ : A (ω ) −→ A (ω ) is compact if and only if, for each δ ∈ (0, mτ ) small enough, one has

2( 1 − 1 ) lim sup τ(a) q p µδ(a) = 0. (4.8) − b r→1 |a|>r

Proof. First we assume that Tµ is compact. Following the proof of the boundedness part 2(p−1) −1 1/2 p and the fact that kKakAp(ωp/2) ω(a) τ(a) , we get the estimate

1/2 2( 1 − 1 ) ω(a) τ(a) q p µδ(a) kTµKak q q/2 kTµkp,ak q q/2 , (4.9) b . 2( 1 −1) A (ω ) . A (ω ) τ(a) p

p p/2 where kp,a are the normalized reproducing kernels in A (ω ). Since, by Lemma 4.16, kp,a tends to zero weakly, and Tµ is compact, the result follows.

p p/2 Conversely, we suppose that (4.8) holds. Let {fn} ⊂ A (ω ) be a bounded sequence converging to zero uniformly on compact subsets of D. By (4.4), we have Z q kTµfnkAq(ωq/2) . |fn(z)| dν(z) = kIνfnkLq(D,dν), (4.10) D

p p/2 q q/2 2(q−1)( 1 − 1 ) where Iν : A (ω ) −→ L (D, dν) with dν(ξ) = ω(ξ) τ(ξ) p q dµ(ξ). By using τ(a) τ(ξ), for ξ ∈ D(δτ(a)), we have Z 1 −q/2 2( 1 − 1 ) sup ω(ξ) dν(ξ) sup τ(a) q p µ (a). 2q/p . bδ |a|>r τ(a) D(δτ(a)) |a|>r

By Theorem E, Iν is compact, and in view of (4.10), Tµ is compact.

Theorem 4.18. Let ω ∈ E, 1 ≤ q < p < ∞ and µ be a ﬁnite positive Borel measure on D satisfy (2.1). The following conditions are equivalent:

p p/2 q q/2 (i) The Toeplitz operator Tµ : A (ω ) −→ A (ω ) is compact.

72 (ii) For each suﬃciently small δ > 0

pq µbδ ∈ L p−q (D, dA). (4.11)

Proof. If Tµ is compact, then it is bounded, and by Theorem 4.15 we get the desired result. Conversely, if (4.11) holds then, by Theorem 4.15, Tµ is bounded. Since, by Theorem 4.12 the spaces Ap(ωp/2) and Aq(ωq/2) are isomorphic to `p, the result is a consequence of a general result of Banach space theory: it is known that, for 1 ≤ q < p < ∞, every bounded operator from `p to `q is compact (see [37, Theorem I.2.7]).

4.6 Hankel operators

One of the most important classes of operators acting on spaces of analytic functions are the Hankel operators. When acting on the classical Hardy spaces, their study presents [50, 54] a broad range of applications such as to control theory, approximation theory, prediction theory, perturbation theory and interpolation problems. Furthermore, one can ﬁnd an extensive literature on Hankel operators acting on other classical function spaces in one or several complex variables, such as Bergman spaces [2, 3, 32, 73, 74], Fock spaces [55] or Dirichlet spaces [67, 68]. In this section, we are going to study big Hankel operators acting on our large weighted Bergman spaces.

Deﬁnition 4.1. Let Mg denote the multiplication operator induced by a function g, and Pω be the Bergman projection, where ω is a weight in the class E. The Hankel operator Hg is given by ω Hg = Hg := (I − Pω)Mg. We assume that the function g satisﬁes

1 1/2 gKz ∈ L (ω ), z ∈ D. (4.12)

ω Under this assumption, the Hankel operator Hg is well-defined on the set E of all finite linear combinations of reproducing kernels and therefore, is densely defined in the weighted Bergman space Ap(ωp/2), 1 ≤ p < ∞. Also, for f ∈ E, one has Z ω Hg f(z) = g(z) − g(s) f(s) Kz(s) ω(s) dA(s). D We are going to study the boundedness and compactness when the symbol is conjugate analytic. In the Hilbert space case A2(ω), and for weights in the class W, a characterization of the boundedness, compactness and membership in Schatten classes of the Hankel 2 2 operator Hg¯ : A (ω) → L (ω) was obtained in [23]. In order to extend such results to the non-Hilbert space setting, we need estimates for the p-norm of the reproducing kernels, and it is here when the condition (4.1) and the exponential type class E enters in action. Before going to study the boundedness of the Hankel operator on Ap(ωp/2) with conjugate analytic symbols we need the following Lemma.

73 Lemma F. Let 1 ≤ p < ∞, g ∈ H(D) and a ∈ D. Then Z 1/p 0 1 p τ(a)|g (a)| ≤ C 2 |g(z) − g(a)| dA(z) . τ(a) D(δτ(a)) Proof. See for example [23]. Now we are ready to characterize the boundedness of the Hankel operator with conjugate analytic symbols acting on large weighted Bergman spaces in term of the growth of the maximum modulus of g0. We begin with the case 1 ≤ p ≤ q < ∞. Theorem 4.19. Let ω ∈ E, 1 ≤ p ≤ q < ∞ and g ∈ H(D) satisfying (4.12). The Hankel p p/2 q q/2 operator Hg : A (ω ) −→ L (ω ) is bounded if and only if

1+2( 1 − 1 ) 0 sup τ(z) q p |g (z)| < ∞. (4.13) z∈D p p/2 q q/2 Proof. Suppose ﬁrst that Hg : A (ω ) −→ L (ω ) is bounded. Thus

kHgKakLq(ωq/2) ≤ kHgk kKakAp (ωp/2). For each z ∈ D, consider the function gz(ξ) := g(z) − g(ξ) Kz(ξ). The condition (4.12) 1 1/2 ensures that gz ∈ A (ω ), and by the reproducing formula in Corollary 4.5, one has

Z HgKa(z) = g(z) − g(ξ) Ka(ξ) Kz(ξ) ω(ξ) dA(ξ) D

= hgz,Kaiω = gz(a).

Because of the boundedness of the Hankel operator Hg, we have q kHgKak q q L (ωq/2) kHgk ≥ q kKakAp(ωp/2) kK kq Z −q a A2(ω) q & kKakAp(ωp/2) q |g(z) − g(a)| dA(z). τ(a) D(δτ(a))

74 Finally, by the estimates on the norm of Kz in Lemma 4.2 and Lemma D, we obtain

1/q 1 1 1 Z 2( q − p ) q kHgk & τ(a) 2 |g(z) − g(a)| dA(z) . τ(a) D(δτ(a))

By Lemma F, this completes the proof of this implication.

Conversely, assume that (4.13) holds and let 1 ≤ p ≤ q < ∞. By Theorem 1.4, there exists a solution u of the equation ∂u = f g0 in Lq(ωq/2) such that Z q q q/2 q kukLq(ωq/2) . |∂u(z)| ω(z) τ(z) dA(z). D

Since any solution v of the ∂-equation has the form v = u−h, with h ∈ H(D), and because Hgf is also a solution of ∂-equation, there is a function h ∈ H(D) such that Hgf = u − h. As a result of Pω Hgf = 0, we have Hgf = I − Pω u, where I is the identity operator. q q/2 Therefore, by the boundedness of Pω on L (ω ) (see Theorem 4.4), we obtain

q q q q kHgfkLq(ωq/2) ≤ kI − Pωk kukLq(ωq/2) . kukLq(ωq/2) (4.14) Z q 0 q q/2 q . |f(z)| |g (z)| ω(z) τ(z) dA(z). D By our assumption (4.13), we have

Z 1 1 q q q/2 2q( p − q ) kHgfkLq(ωq/2) . |f(z)| ω(z) τ(z) dA(z). D On the other hand, by Lemma A

1/2 −2/p |f(z)| ω(z) . τ(z) kfkAp(ωp/2). Using the last pointwise estimate, we have Z q q−p p p/2 q kHgfkLq(ωq/2) . kfkAp(ωp/2) |f(z)| ω(z) dA(z) = kfkAp(ωp/2). D This completes the proof.

Next we are going to characterize the boundedness of the Hankel operator with conjugate analytic symbols when 1 ≤ q < p < ∞. Before that we prove the following Lemma:

75 Lemma 4.20. Let δ0 ∈ (0, mτ ) and 0 < r < ∞. Then Z 0 r 1 r |f (z)| . r+2 |f(s)| dA(s), τ(z) D(δ0τ(z))

for f ∈ H(D). Proof. By Cauchy’s integral formula and Lemma A, we get Z Z 0 f(η) 1 |f (z)| . 2 dη . 2 |f(η)| |dη| δ0τ(z) (η − z) τ(z) δ0τ(z) |η−z|= 4 |η−z|= 4 Z Z !1/r 1 1 r . 2 2 |f(s)| dA(s) |dη|. τ(z) δ0τ(z) τ(η) |η−z|= 4 D(δ0τ(η)/4)

An application of τ(η) τ(z), for η ∈ D(δ0τ(z)/2), gives

Z Z !1/r 0 r 1 1 r |f (z)| . 2 2 |f(s)| dA(s) |dη| τ(z) δ0τ(z) τ(z) |η−z|= 4 D(δ0τ(z)/2) !1/r 1 Z |f(s)|rdA(s) , . 1+ 2 τ(z) r D(δ0τ(z)/2)

which proves the desired result.

The following result gives the characterization of the boundedness of the Hankel operator going from Ap(ωp/2) into Lq(ωq/2) when 1 ≤ q < p < ∞.

Theorem 4.21. Let ω ∈ E, 1 ≤ q < p < ∞ and g ∈ H(D) satisfying (4.12). Then the following statement are equivalent:

p p/2 q q/2 (a) The Hankel operator Hg : A (ω ) −→ L (ω ) is bounded.

0 r 1 1 1 (b) The function τg belongs to L (D, dA), where r = q − p .

Proof. Suppose that τ g0 ∈ Lr(D, dA). By (4.14), since p/q > 1 a simple application of H¨older’sinequality, we get Z kH fkq |f(z)|q |g0(z)|q ω(z)q/2 τ(z)q dA(z) ≤ kfkq kτ g0kq . g Lq(ωq/2) . Ap(ωp/2) Lr(D,dA) D

This proves the boundedness of Hg.

76 p Conversely, pick ε > 0 and let {zk} be an (ε, τ)- lattice on D. For a sequence λ = {λk} ∈ ` , we consider the function ∞ p−1 X 1/2 2( p ) Gt(z) = λk rk(t) ω(zk) τ(zk) Kzk (z) , 0 < t < 1, k=0 where rk(t) is a sequence of Rademacher functions. Because of the norm estimate for reproducing kernels given in Lemma 4.2, by part (i) of Theorem 4.12, we obtain

kGtkAp(ωp/2) . kλk`p .

Thus, the boundedness of Hg gives

q q kHg GtkLq(ωq/2) . kλk`p . Therefore,

∞ Z p−1 q X 1/2 2( p ) q/2 q λ r (t) ω(z ) τ(z ) H K (z) ω(z) dA(z) kλk p . k k k k g zk . ` D k=0 Using the same method in (4.6), we obtain

∞ Z p−1 q X q q/2 2q( p ) q/2 q |λk| ω(zk) τ(zk) HgKzk (z) χk(z) ω(z) dA(z) . kλk`p , (4.15) D k=0 which χk is the characteristic function of the set D(3 ε τ(zk)). Additionally, by Lemma E, Lemma D and (1.2), we get

ω(z )−q/2 H K (z) q ω(z)q/2 = g(z) − g(z ) q |K (z)|q ω(z)q/2 k g(z) − g(z ) q. g zk k zk & 2q k τ(zk) Putting this in (4.15), it gives ∞ Z X q −2q/p q q |λk| τ(zk) g(z) − g(zk) dA(z) . kλk`p . k=0 D(3 ε τ(zk)) Furthermore, by Lemma F we obtain

∞ X q −2q/p+2 0 q q |λk| τ(zk) |τ(zk)g (zk)| . kλk`p . k=0

p Moreover, by the duality between `p/q and ` p−q , it follows

∞ X 2 0 r Ig0 := τ(zk) τ(zk)|g (zk)| < ∞. k=0

77 0 r In order to ﬁnish the proof we will justify that kτ g k I 0 . For that, on the one Lr(D,dA) . g hand by Lemma 4.20 applied to g0, we get Z 00 r 1 0 r g (ξ) . r+2 |g (s)| dA(s). (4.16) τ(ξ) D(δ0τ(ξ))

On the other hand, by Cauchy estimates there exists ξ ∈ [z, zk] such that 0 0 00 g (z) − g (zk) ≤ |g (ξ)||z − zk|.

Using τ(ξ) τ(z) τ(zk), for ξ, z ∈ D(ε τ(zk)), we have

Z 0 r X r 0 r kτ g kLr(D,dA) ≤ τ(z) |g (z)| dA(z) k D(ε τ(zk)) Z X r 0 0 r 2 ≤ C τ(zk) g (z) − g (zk) dA(z) + Cε Ig0 k D(ε τ(zk)) Z r X 2r 00 r 2 ≤ Cε τ(zk) g (ξ) dA(z) + Cε Ig0 . k D(ε τ(zk)) By (4.16) and using again (1.2), we obtain

Z 2r Z 0 r r X τ(zk) 0 r 2 kτ g k r ≤ C ε |g (s)| dA(s) dA(z) + ε I 0 L (D,dA) τ(ξ)r+2 g k D(ε τ(zk)) D(δ0τ(ξ)) Z r+2 X r 0 r 2 ≤ C ε τ(s) |g (s)| dA(s) + ε Ig0 . k D(3δ0τ(zk)) −2 By Lemma B, every point z ∈ D belongs to at the most Cε of the sets D(3δ0τ(zk)). Hence

r 0 r 0 1 − C ε kτ g kLr(D,dA) . Ig . Thus, taking ε so that C εr < 1/2, we get the desired result.

Next we characterize the compactness of the Hankel operator with conjugate analytic symbol acting from Ap(ωp/2) into Lq(ωq/2), 1 ≤ p, q < ∞. This characterization will be given in two theorems depending on the order of p and q. We begin with the case 1 ≤ p ≤ q < ∞.

Theorem 4.22. Let ω ∈ E, 1 < p ≤ q < ∞ and g ∈ H(D) satisfying (4.12). Then, the p p/2 q q/2 Hankel operator Hg : A (ω ) −→ L (ω ) is compact if and only if

1+2( 1 − 1 ) 0 lim τ(a) q p |g (a)| = 0. |a|→1−

78 Proof. Let a ∈ D and 1 < p ≤ q < ∞. Recall that ka,p is the normalized reproducing kernel p p/2 in A (ω ). By Lemma 4.16, ka,p → 0 weakly. Thus, if Hg is compact, then

lim kHg ka,pkLq(ωq/2) = 0. |a|→1−

Let δ be small enough such that |Ka(z)| kKakA2(ω) kKzkA2(ω), for z ∈ D(δτ(a)). By 1−p −1/2 2( p ) Lemma D, using (1.2) and the norm estimate kKakAp(ωp/2) ω(a) τ(a) , we have

Z q q/2 q q |Ka(z)| ω(z) kHg ka,pkLq(ωq/2) ≥ g(a) − g(z) q dA(z) D(δτ(a)) kKakAp(ωp/2) Z 1 q & 2q/p g(a) − g(z) dA(z). τ(a) D(δτ(a)) It follows from Lemma F,

1 1 1+2( q − p ) 0 kHg ka,pkLq(ωq/2) & τ(a) |g (a)|. This implies that 1+2( 1 − 1 ) 0 lim τ(a) q p |g (a)| = 0, |a|→1− which completes the proof of this implication.

p p/2 Conversely, let {fn} be a bounded sequence in A (ω ) such that fn → 0 uniformly on compact subsets of D. To show compactness, it is standard to see that it is enough to prove that kHg¯fnkLq(ωq/2) → 0. By the assumption, given any ε > 0, there is 0 < r0 < 1 such that 1+2( 1 − 1 ) 0 τ(z) q p |g (z)| < ε, r0 < |z| < 1.

Since {fn} converges to zero uniformly on compact subsets of D, exists an integer n0 such that |fn(z)| < ε for |z| ≤ r0 and n ≥ n0. According to (4.14), we have Z q q 0 q q/2 q kHgfnkLq(ωq/2) . |fn(z)| |g (z)| ω(z) τ(z) dA(z) D

Z Z q 0 q q/2 q = + |fn(z)| |g (z)| ω(z) τ(z) dA(z). |z|≤r0 r0<|z|<1 On the one hand, it is easy to see that Z q 0 q q/2 q q |fn(z)| |g (z)| ω(z) τ(z) dA(z) . ε . (4.17) |z|≤r0

79 On the other hand, by Lemma A, we have the pointwise estimate

−1/2 −2/p |fn(z)| . ω(z) τ(z) kfnkAp(ωp/2). Applying this together with our assumption, we get Z q 0 q q/2 q |fn(z)| |g (z)| ω(z) τ(z) dA(z) r0<|z|<1

Z q p q−p q/2 2q 1 − 1 < ε |fn(z)| |fn(z)| ω(z) τ(z) p q dA(z) D

Z q q−p p p/2 q q . ε kfnkAp(ωp/2) |fn(z)| ω(z) dA(z) = ε kfnkAp(ωp/2). D

Combining this with (4.17) gives us that lim kHgfnkLq(ωq/2) = 0. This shows that the n→∞ p p/2 q q/2 Hankel operator Hg : A (ω ) −→ L (ω ) is compact.

The next Theorem give us the compactness characterization of the Hankel operator when 1 ≤ q < p < ∞.

Theorem 4.23. Let ω ∈ E, 1 ≤ q < p < ∞ and g ∈ H(D) satisfying (4.12). The following conditions are equivalent:

p p/2 q q/2 (a) The Hankel operator Hg : A (ω ) −→ L (ω ) is compact.

0 r 1 1 1 (b) The function τg belongs to L (D, dA), where r = q − p .

Proof. (a) =⇒ (b) Assume that Hg is compact. Then Hg is bounded. Hence, by applying Theorem 4.21 we get the desired result. 0 r 1 1 1 (b) =⇒ (a) Suppose that τg belongs to L (D, dA), where r = q − p . By Theorem 4.21, p p/2 the Hankel operator Hg is bounded, and as a result of Theorem 4.12 the space A (ω ) is p isomorphic to ` . In this case, Hg is also compact, due to a general result of Banach space theory: it is known that, for 1 ≤ q < p < ∞, every bounded operator from `p to `q is compact (see [37, Theorem I.2.7]). This ﬁnishes the proof.

4.7 Examples of weights in the class E

In this Section, we are going to show that the family of exponential type weights ωσ given by −A ω (z) = exp , σ > 0, A > 0. (4.18) σ (1 − |z|2)σ

80 satisfy condition (4.1), and therefore they are in the class E. Notice that the case σ = 1 was obtained in [16], and we follow his method with appropriate modiﬁcations.

For λ > 0, let Z 1 λ vλ = r ωσ(r) dr 0 Because the functions zn en(z) = √ 2v2n+1 2 2 form an orthonormal basis of A (ωσ), the reproducing kernel Kz of A (ωσ) is given by X (¯zw)n K (w) = . z 2v n 2n+1

We let K(z) = P zn . We need the following result. n 2v2n+1 Proposition 4.24. Let 0 < σ < ∞. Then

Z 2π 1+ σ iθ dθ 1 2 A M1(r, K) := |K(re )| . exp σ 0 2π 1 − r (1 − r)

We postpone the proof of the proposition for a moment, and we ﬁrst use it in order to show that the exponential weights are in the class E.

81 Lemma 4.26. For 0 < σ < ∞. There is a positive constant C depending on σ such that

s − r r s g (s) ≤ C − . r 1 − sr (1 − r2)σ (1 − s2)σ

Proof. We may assume that r ≤ s, because changing the order of r and s does not aﬀect the result. Observe that

1 1 1 1 1 1 g (s) = − + − . (4.19) r 2 (1 − rs)β (1 − r2)β 2 (1 − rs)β (1 − s2)β

By the mean value theorem of diﬀerential calculus, there exists x1 ∈ (r, s) such that

2 σ σ σ−1 (1 − r ) − (1 − rs) = σr (1 − rx1) (s − r).

σ−1 We ﬁrst consider the case 0 < σ ≤ 1. As σ ≤ 1 and x1 ≤ s, we have (1 − rx1) ≤ (1 − rs)σ−1. Then

1 1 (1 − r2)σ − (1 − rs)σ − = (1 − rs)σ (1 − r2)σ (1 − r2)σ(1 − rs)σ (4.20) σr (s − r) ≤ . (1 − r2)σ(1 − rs)

Similarly, there exists x2 ∈ (r, s) such that

1 1 −σs (1 − sx )σ−1(s − r) − = 2 . (1 − rs)σ (1 − s2)σ (1 − s2)σ(1 − rs)σ

Since (1 − sx2) < (1 − rs) and σ ≤ 1, we get 1 1 s (s − r) − ≤ −σ . (1 − rs)σ (1 − s2)σ (1 − s2)σ(1 − rs)

Combining this one with (4.19) and bearing in mind (4.20), we get the desired result for 0 < σ ≤ 1 with C = σ/2.

Next, we consider the case 1 < σ ≤ 2. In this case, as σ/2 ≤ 1, arguing as before we get

(1 − r2)σ − (1 − rs)σ = ((1 − r2)σ/2 − (1 − rs)σ/2) · ((1 − r2)σ/2 + (1 − rs)σ/2)

σ ≤ r (1 − rs)σ/2−1(s − r) (1 − r2)σ/2 + (1 − rs)σ/2 . 2 82 This gives

1 1 σ r(s − r) σ r(s − r) − ≤ + . (4.21) (1 − rs)σ (1 − r2)σ 2 (1 − r2)σ(1 − rs) 2 (1 − r2)σ/2(1 − rs)σ/2+1

Similarly, σ (1 − s2)σ − (1 − rs)σ ≤ − s(1 − rs)σ/2−1(s − r) (1 − s2)σ/2 + (1 − rs)σ/2 , 2 and we obtain 1 1 σ s(s − r) σ s(s − r) − ≤ − − . (4.22) (1 − rs)σ (1 − s2)σ 2 (1 − s2)σ(1 − rs) 2 (1 − s2)σ/2(1 − rs)σ/2+1

Inserting (4.21) and (4.22) into (4.19) we get

σ s − r r s g (s) ≤ − r 4 1 − rs (1 − r2)σ (1 − s2)σ

σ s − r r s + σ +1 2 σ/2 − 2 σ/2 . 4 (1 − rs) 2 (1 − r ) (1 − s )

Hence the result follows in this case with C = σ/4, because every summand is negative.

Next, we show by induction that, for 2m−1 < σ ≤ 2m with m ≥ 1, we have

σ s − r r s gσ(s) = g (s) ≤ − . r r 2m 1 − rs (1 − r2)σ (1 − s2)σ

We have just proved the case m = 1. So assume m ≥ 2 and that the result is true for m − 1 and proceed to prove the case m. Using the identities (1 − r2)σ − (1 − rs)σ = (1 − r2)σ/2 − (1 − rs)σ/2 · (1 − r2)σ/2 + (1 − rs)σ/2 and (1 − s2)σ − (1 − rs)σ = (1 − s2)σ/2 − (1 − rs)σ/2 · (1 − s2)σ/2 + (1 − rs)σ/2 we get

1 1 1 1 − = (1 − r2)−σ/2 − (1 − rs)σ (1 − r2)σ (1 − rs)σ/2 (1 − r2)σ/2 (4.23) 1 1 + (1 − rs)−σ/2 − (1 − rs)σ/2 (1 − r2)σ/2

83 and 1 1 1 1 − = (1 − s2)−σ/2 − (1 − rs)σ (1 − s2)σ (1 − rs)σ/2 (1 − s2)σ/2 (4.24) 1 1 + (1 − rs)−σ/2 − . (1 − rs)σ/2 (1 − s2)σ/2

Adding the two identities, we obtain

1 1 2gσ(s) = (1 − r2)−σ/2 − r (1 − rs)σ/2 (1 − r2)σ/2 1 1 + (1 − s2)−σ/2 − + (1 − rs)−σ/2 2gσ/2(s). (1 − rs)σ/2 (1 − s2)σ/2 r

σ/2 As gr (s) ≤ 0, this gives

1 1 1 1 2gσ(s) ≤ (1−r2)−σ/2 − +(1−s2)−σ/2 − . r (1 − rs)σ/2 (1 − r2)σ/2 (1 − rs)σ/2 (1 − s2)σ/2

Use again the identities (4.23) and (4.24), but with σ replaced by σ/2 in order to get

σ 2gr (s) ≤ (I) + (II) + (III) + (IV ), with 1 1 (I) := (1 − r2)−σ/2(1 − r2)−σ/4 − , (1 − rs)σ/4 (1 − r2)σ/4

1 1 (II) := (1 − r2)−σ/2(1 − rs)−σ/4 − (1 − rs)σ/4 (1 − r2)σ/4

1 1 (III) := (1 − s2)−σ/2(1 − s2)−σ/4 − (1 − rs)σ/4 (1 − s2)σ/4 and 1 1 (IV ) := (1 − s2)−σ/2(1 − rs)−σ/4 − . (1 − rs)σ/4 (1 − s2)σ/4 Note that, as r ≤ s, the term in brackets in (II) is positive, and also we have (1−r2)−σ/2 ≤ (1 − s2)−σ/2. Therefore

2 −σ/2 −σ/4 σ/4 (II) + (IV ) ≤ 2(1 − s ) (1 − rs) gr (s) ≤ 0.

84 This gives

1 1 2gσ(s) ≤ (I) + (III) = (1 − r2)−σ/2(1 − r2)−σ/4 − r (1 − rs)σ/4 (1 − r2)σ/4

1 1 + (1 − s2)−σ/2(1 − s2)−σ/4 − (1 − rs)σ/4 (1 − s2)σ/4

(1 − r2)σ/4 − (1 − rs)σ/4 (1 − s2)σ/4 − (1 − rs)σ/4 = + . (1 − rs)σ/4 (1 − r2)σ (1 − rs)σ/4 (1 − s2)σ

Iterating this process, we arrive at

(1 − r2)σ/2m − (1 − rs)σ/2m (1 − s2)σ/2m − (1 − rs)σ/2m 2gσ(s) ≤ + . r (1 − rs)σ/2m (1 − r2)σ (1 − rs)σ/2m (1 − s2)σ

As σ/2m ≤ 1, using the meanvalue theorem as in the case σ ≤ 1, we have

(1 − r2)σ/2m − (1 − rs)σ/2m ≤ σ/2mr(1 − rs)σ/2m−1 (s − r),

and (1 − s2)σ/2m − (1 − rs)σ/2m ≤ −σ/2ms(1 − rs)σ/2m−1 (s − r). This ﬁnally gives

σ s − r r s 2gσ(s) ≤ − . r 2m 1 − rs (1 − r2)σ (1 − s2)σ

Now, we can continue with the proof. Set

1 fr(s) = exp Agr(s) 1+ σ . (1 − rs) 2

We must show that Z 1 fr(s) ds ≤ C (4.25) 0 for some constant C not depending on r. Without loss of generality, we may assume that r > 1/2. From Lemma 4.26, we see that gr(s) ≤ 0. Hence,

Z 1/2 Z 1/2 ds σ/2 fr(s) ds ≤ 1+ σ ≤ 2 . (4.26) 0 0 (1 − rs) 2

85 Also,

σ σ Z r+(1−r)1+ 2 Z r+(1−r)1+ 2 ds σ/2 f (s) ds ≤ σ ≤ 2 σ r σ 1+ r−(1−r)1+ 2 r−(1−r)1+ 2 (1 − rs) 2 (4.27)

1+ σ Z r+(1−r) 2 ds ≤ σ ≤ 2. σ 1+ r−(1−r)1+ 2 (1 − r) 2

In view of (4.26) and (4.27), to see that (4.25) holds, it remains to prove that

Z 1 f (s) ds ≤ C (4.28) σ r 1 r+(1−r)1+ 2

and σ Z r−(1−r)1+ 2 fr(s) ds ≤ C2. (4.29) 1/2 We begin with the proof of (4.28). Let N = N(r) be the largest positive integer such that 1 2N < . (1 − r)σ/2 Observe that k 1+ σ k −σ/2 r + 2 (1 − r) 2 ≥ 1 ⇔ 2 ≥ (1 − r) . Then N Z 1 X Z fr(s) ds = fr(s) ds, 1+σ/2 r+(1−r) k=0 Jk(r)∩[0,1] with k 1+ σ k+1 1+ σ Jk(r) = r + 2 (1 − r) 2 , r + 2 (1 − r) 2 . As s > r, from Lemma 4.26 we have

gr(s) ≤ −CAr(s) · Br(s) (4.30)

with s − r A (s) := r 1 − sr and s r B (s) := − . r (1 − s2)σ (1 − r2)σ

For s ∈ Jk(r), we have s − r A (s) ≥ ≥ 2k−1(1 − r)σ/2. (4.31) r 2(1 − r)

86 Let h(x) = x/(1 − x2)σ. Then

1 − x2 + 2σx2 h0(x) = . (1 − x2)σ+1

By the mean value theorem, there is x ∈ (r, s) with

1 + (2σ − 1)x2 B (s) = (s − r). r (1 − x2)σ+1

If 2σ ≥ 1, then 1+(2σ−1)x2 ≥ 1 and, if 2σ < 1, because x < 1, we have 1+(2σ−1)x2 > 2σ. Therefore, 2 1 + (2σ − 1)x ≥ Cσ := min(1, 2σ). Hence, as x > r, we obtain s − r B (s) ≥ C . r σ (1 − r2)σ+1 Since k 1+ σ s − r ≥ 2 (1 − r) 2 , s ∈ Jk(r), we have k −σ/2 Br(s) ≥ C(σ)2 (1 − r) . Putting this estimate together with (4.31) into (4.30) we obtain

2k gr(s) ≤ −C(σ)2 , s ∈ Jk(r).

Hence,

1 N Z Z σ X Agr(s) −1− fr(s) ds = e (1 − sr) 2 ds 1+σ/2 r+(1−r) k=0 Jk(r)∩[0,1]

N Z X −C22k ds ≤ e 1+ σ (1 − sr) 2 k=0 Jk(r)

N N X −C22k |Jk(r)| X k −C22k ≤ e 1+ σ = 2 e = C1 < ∞. (1 − r) 2 k=0 k=0 This proves (4.28). Therefore, in order to ﬁnish the proof of this case, it remains to prove that (4.29) holds, that is, we must show that

σ Z r−(1−r)1+ 2 fr(s) ds ≤ C2. 1/2

87 Let M = M(r) be the largest positive integer such that

r − 1/2 2M < . (1 − r)1+β/2

Then

1+σ/2 M Z r−(1−r) X Z fr(s) ds ≤ fr(s) ds (4.32) 1/2 k=0 Jk(r)∩[1/2,1] with k+1 1+σ/2 k 1+σ/2 Jk(r) = r − 2 (1 − r) , r − 2 (1 − r) . Now r > s, so that from Lemma 4.26 we get

gr(s) ≤ −C(σ) Cr(s) · Dr(s) (4.33) with r − s r s C (s) := , and D (s) := − . r 1 − sr r (1 − r2)σ (1 − s2)σ

We have k 1+σ/2 r − s ≥ 2 (1 − r) , s ∈ Jk(r), and

k+1 1+ σ k σ 1 − sr ≤ 1 − (r − 2 (1 − r) 2 )r ≤ 2(1 − r) 1 + 2 (1 − r) 2 , s ∈ Jk(r).

This gives 2k−1(1 − r)σ/2 C (s) ≥ , s ∈ J (r). (4.34) r 1 + 2k(1 − r)σ/2 k

m In order to estimate the term Dr(s), take a non-negative integer m with 2 σ ≥ 1. Then we claim that

r(1 − s2)2mσ − s(1 − r2)2mσ D (s) ≥ 2−m . (4.35) r (1 − r2)σ(1 − s2)2mσ We prove the claim by induction on m. The result is obvious for m = 0. Assume the claim is true for m and proceed to show that the case m + 1 also holds. By induction, we have

r(1 − s2)2mσ − s(1 − r2)2mσ D (s) ≥ 2−m r (1 − r2)σ(1 − s2)2mσ

r2(1 − s2)2m+1σ − s2(1 − r2)2m+1σ = 2−m . (1 − r2)σ(1 − s2)2mσr(1 − s2)2mσ + s(1 − r2)2mσ

88 Because s ≤ r, then (1 − r2) ≤ (1 − s2) and

2−m r r(1 − s2)2m+1σ − s(1 − r2)2m+1σ D (s) ≥ · r r + s (1 − r2)σ(1 − s2)2m+1σ

r(1 − s2)2m+1σ − s(1 − r2)2m+1σ ≥ 2−(m+1) · . (1 − r2)σ(1 − s2)2m+1σ Hence the claim is proved.

As 2mσ ≥ 1 and s ≤ r, then (1 − r2)2mσ−1 ≤ (1 − s2)2mσ−1, and using the inequality (4.35), we obtain r(1 − s2)2mσ − s(1 − r2)2mσ−1(1 − r2) D (s) ≥ 2−m r (1 − r2)σ(1 − s2)2mσ

r(1 − s2) − s(1 − r2) r − s ≥ 2−m ≥ 2−m . (1 − r2)σ(1 − s2) (1 − r2)σ(1 − s2)

Now, for s ∈ Jk(r), we have r − s ≥ 2k(1 − r)1+σ/2, and 1 − s2 ≤ 2(1 − s) ≤ 21 − (r − 2k+1(1 − r)1+σ/2) = 2(1 − r)(1 + 2k+1(1 − r)σ/2). This gives 2−(m+1) 2k(1 − r)−σ/2 2k(1 − r)−σ/2 D (s) ≥ ≥ 2−(m+1+σ) · , s ∈ J (r). r (1 + r)σ(1 + 2k+1(1 − r)σ/2) (1 + 2k+1(1 − r)σ/2) k Putting this together with (4.34) into (4.33) we obtain 22k g (s) ≤ −C(σ) · , s ∈ J (r); 1/2 ≤ s < 1. r (1 + 2k+1(1 − r)σ/2)2 k Then

r−(1−r)1+σ/2 M Z Z σ X Agr(s) −1− fr(s) ds ≤ e (1 − sr) 2 ds 1/2 k=0 Jk(r)∩[1/2,1] M X 22k Z ds ≤ exp − C 2 1+ σ ds. k+1 σ/2 (1 − sr) 2 k=0 1 + 2 (1 − r) Jk(r)∩[1/2,1] Since

k 1+σ/2 k σ/2 1 − sr ≥ 1 − s ≥ 1 − (r − 2 (1 − r) ) = (1 − r)(1 + 2 (1 − r) ), s ∈ Jk(r).

89 k 1+σ/2 and |Jk(r)| = 2 (1 − r) we get

1+σ/2 M Z r−(1−r) X 22k 2k f (s) ds ≤ exp − C . (4.36) r 2 σ 1+ σ k+1 σ/2 k 2 1/2 k=0 1 + 2 (1 − r) 1 + 2 (1 − r) 2

σ/2 −(1+σ/2) Because hr(x) = (1 + x(1 − r) ) is a decreasing function, then

k 2k Z 2 dx ≤ 2 . k σ 1+σ/2 σ 1+σ/2 (1 + 2 (1 − r) 2 ) 2k−1 (1 + x(1 − r) 2 )

Also, for x ∈ [2k−1, 2k], we have

22k 22k x2 ≥ ≥ . (1 + 2k+1(1 − r)σ/2)2 16(1 + 2k−1(1 − r)σ/2)2 16(1 + x(1 − r)σ/2)2 Therefore, we obtain

M X 22k 2k exp − C (1 + 2k+1(1 − r)σ/2)2 k σ/21+σ/2 k=0 1 + 2 (1 − r)

−(1+ σ ) Z (1−r) 2 −Cx2 dx ≤ 2 exp . 2 σ/2 1+σ/2 1/2 1 + x(1 − r)σ/2 (1 + x(1 − r) )

Bearing in mind (4.36), we arrive at

σ 1+σ/2 −(1+ ) Z r−(1−r) Z (1−r) 2 −Cx2 dx f (s) ds ≤ 2 exp . (4.37) r 2 σ/2 1+σ/2 1/2 1/2 1 + x(1 − r)σ/2 (1 + x(1 − r) )

Consider the function 2 −Cx σ −(1+σ/2) g(x) = exp 1 + x(1 − r) 2 . (1 + x(1 − r)σ/2)2 To ﬁnish the proof it remains to see that

σ σ Z (1−r)− 2 Z (1−r)−(1+ 2 ) g(x) dx ≤ K and g(x) dx ≤ K 1 σ 2 1/2 (1−r)− 2 for some positive constants K1 and K2 not depending on r. It is clear that

σ Z (1−r)− 2 Z ∞ −Cx2/4 g(x) dx ≤ e dx = K1 < ∞. 1/2 1/2

90 Finally, using the change of variables t = (1 − r)σ/2x and the fact that the function k(t) = −t(1 + t)−1 decreases, we get

σ −(1+ ) −1 Z (1−r) 2 Z (1−r) −Ct2 dt g(x) dx = exp σ σ 2 σ/2 1+σ/2 (1−r)− 2 1 (1 − r) (1 + t) (1 − r) (1 + t)

Z (1−r)−1 −σ/2 −C dt ≤ (1 − r) exp σ 1+σ/2 ≤ K2 4(1 − r) 1 (1 + t)

for some constant K2 not depending on r. This completes the proof.

The rest of the section is devoted to the proof of Proposition 4.24. We need ﬁrst some lemmas, whose proof uses results of [33] involving the Legendre-Fenschel transform.

1 1 − σ Lemma 4.27. Let A, σ > 0 and let c0 = A σ+1 σ σ+1 + σ σ+1 . Then ∞ 1+σ/2 X σ n 1 A exp c n σ+1 ρ exp . 0 . 1 − ρ (1 − ρ)σ n=1 Proof. We have ∞ ∞ X σ n X N(n) n exp c0n σ+1 ρ = e ρ n=1 n=1 with σ/(σ+1) N(x) = c0 x , x > 0. A calculation shows that its inverse Legendre-Fenschel transform is given by u(t) = sup N(x) − xt = A t−σ. x>0 Since u00(t) = Aσ(σ + 1)t−(2+σ), by Corollary 1 in [33] we get

∞ X eN(n) ρn pu00(log 1/ρ) eu(log 1/ρ) n=1

1 1+σ/2 A = pAσ(σ + 1) exp log 1/ρ (log 1/ρ)σ

1 1+σ/2 A ≤ pAσ(σ + 1) exp . 1 − ρ (1 − ρ)σ

91 Lemma 4.28. Let 0 < σ < ∞ and Z 1 A I(λ) = rλ exp − dr, A > 0. 1 σ 0 (log r ) Then − (2+σ) σ I(λ) λ 2(σ+1) exp − c0λ σ+1 , λ → ∞, 1 1 − σ where c0 = A σ+1 σ σ+1 + σ σ+1 . Proof. This is proved for 0 < σ ≤ 1 in [18, 19], but since we need the explicit expression of the constant c0 we give the proof here. After the change of variables t = − log r, we have Z ∞ Z ∞ I(λ) = e−At−σ−(λ+1)tdt = e−v(t)−(λ+1)tdt, 0 0 with v(t) = At−σ. Consider its Legendre-Fenschel transform

L(x) = inf v(t) + xt. t>0

(2+σ) σ/(σ+1) 00 σ − σ+1 A simple calculation shows that L(x) = c0x . Since L (x) = −c0 (σ+1)2 x , by [33, Theorem 1], we have

(2+σ) σ p 00 −L(λ) − I(λ) −L (λ) e λ 2(σ+1) exp − c0λ σ+1 , λ → ∞.

Proof of Proposition 4.24

Step 1. Following the same argument as in [16] we get

∞ X M1(ρ, K) ≤ C + Mn, n=3 with ρx Mn := max . x∈[2n−1,2n+1] v2x+1 We recall that Z 1 λ vλ = s ωσ(s) ds. 0

Step 2. Changing variables and then using Lemma 4.28 we have

−(2+σ) σ v2x+1 x 2σ+2 exp − c0 x σ+1 , x → ∞,

92 Therefore, x ρ x 2+σ σ h(x) ρ x 2σ+2 exp c0 x σ+1 = e , x → ∞, v2x+1 where 2 + σ σ h(x) = x log ρ + log x + c x σ+1 . 2σ + 2 0

(2+σ) (σ+2) 00 σ − σ+1 Since h (x) = − 2(σ+1)x2 − (σ+1)2 x < 0, then h can only have a critical point xρ, and in case it exists, it must be a maximum. Next we are going to show that it has a unique 1 σ+1 σ+1 σ+1 maximum xρ which is comparable with yρ = ( 1 ) . Indeed, for n ≥ σc we have log ρ 0

σ+1 0 1 (2 + σ)n 1 σ c0 1 1 h y = log ρ + log + n σ+1 log n ρ 2(σ + 1) ρ σ + 1 ρ

σ+1 1 σ c0 1 (2 + σ)n 1 = log · n σ+1 − 1 + log > 0. ρ σ + 1 2(σ + 1) ρ

1 σ+1 2+σ(1+2c0) Also, for m ≥ max (log ρ ) ; 2(σ+1) , (note that we can take m independent of ρ, 1 σ+1 since (log ρ ) is bounded for 1/2 ≤ ρ < 1)

0 2 + σ(1 + 2c0) −1 h (m y ) ≤ log ρ + (m y ) σ+1 ρ 2(σ + 1) ρ

! 1 2 + σ(1 + 2c0) = log 1 − 1 < 0. ρ 2(σ + 1)m σ+1

1 In the last inequality we used that x β+1 ≤ x for x ≥ 1. Therefore, there exists xρ ∈ yρ 0 ( n , myρ) such that h (xρ) = 0. This gives us

1 1+σ/2 h(x) h(xρ) sup e = e exp g(xρ) , (4.38) x∈(0,∞) 1 − ρ

with σ g(x) = x log ρ + c0x σ+1 . now, a calculation shows that g has a unique maximum at the point

!σ+1 σc0 1 zρ = 1 . (σ + 1) log ρ

93 We have !σ σc σ+1 1 σc 1 g(z ) = − 0 + c 0 ρ 1 σ 0 1 σ + 1 (log ρ ) (σ + 1) log ρ

σc σ 1 σc = 0 c − 0 1 σ 0 σ + 1 (log ρ ) σ + 1

c σ+1 1 A = σσ 0 = , 1 σ 1 σ σ + 1 (log ρ ) (log ρ ) 1 1 − σ because, as c0 = A σ+1 σ σ+1 + σ σ+1 , then

σ+1 σ σ+1 σ c0 A σ 1 − σ σ = σ σ+1 + σ σ+1 = A. σ + 1 (σ + 1)σ+1

This together with (4.38) yields

1+σ/2 h(x) 1 sup e . exp g(zρ) x∈(0,∞) 1 − ρ 1 1+σ/2 A = exp (4.39) 1 σ 1 − ρ (log ρ ) 1 1+σ/2 A ≤ exp . 1 − ρ (1 − ρ)σ

n0 n0+1 Step 3. Choose n0 ∈ N such that 2 ≤ xρ < 2 and split the above sum as follows:

∞ n −2 n +1 ∞ X X0 X0 X Mn = Mn + Mn + Mn.

n=1 n=1 n=n0−1 n=n0+2

For 1 ≤ n ≤ n0 − 2, because of the monotonicity of h

x ρ h(x) h(2n+1) Mn = max = max e = e . x∈[2n−1,2n+1] v2x+1 x∈[2n−1,2n+1] And it follows from (4.39) that

n0+1 1+σ/2 X 1 A M exp . n . 1 − ρ (1 − ρ)σ n=n0−1

94 Using again the monotonicity of h for n ≥ n0 + 2,

x ρ h(x) h(2n−1) Mn = max = max e = e . x∈[2n−1,2n+1] v2x+1 x∈[2n−1,2n+1] So,

∞ n0−2 ∞ 1+σ/2 X X n+1 X n−1 1 A M ≤ eh(2 ) + eh(2 ) + exp . (4.40) n 1 − ρ (1 − ρ)σ n=1 n=1 n=n0+2

Next, for n ≤ n0 − 2 and since h increases,

n+2 n+2 n+2 2 −1 2 −1 2 −1 h(k) n+1 1 X n+1 1 X X e eh(2 ) = eh(2 ) ≤ eh(k) ≤ 2 . 2n+1 2n+1 k k=2n+1 k=2n+1 k=2n+1

And for n ≥ n0 + 2 and since h decreases,

n−1 n−1 n−1 2 −1 2 −1 2 −1 h(k) n−1 1 X n−1 4 X X e eh(2 ) = eh(2 ) ≤ eh(k) ≤ 2 . 2n−2 2n k k=2n−2 k=2n−2 k=2n−2 Using this one in (4.40), we have

1+σ/2 ∞ 1 A X eh(k) M (r, K) exp + 1 . 1 − ρ (1 − ρ)σ k k=1

1+σ/2 ∞ 1 A X − σ σ k exp + k 2σ+2 exp c k σ+1 ρ . 1 − ρ (1 − ρ)σ 0 k=1 Finally, as ∞ ∞ X − σ σ k X σ k k 2σ+2 exp c0 k σ+1 ρ ≤ exp c0 k σ+1 ρ , k=1 k=1 applying Lemma 4.27, we get the result. The proof is complete.

95 96 Chapter 5

Open questions and future research

We believe that we have done a satisfactory work in order to get a better understanding of the function properties of large weighted Bergman spaces and the operators acting on them. We hope that this work is going to attract many other researchers to this area, and expect that the study of this function spaces is going to experience a period of intensive research in the next years. However, we have not been able to solve all the problems we had in mind, and in this last chapter we discuss some open problems we left, as well as some other problems we think it can be interesting to look on the future.

5.1 Extension of the results to p =6 2

One can look if it is possible to extend the results obtained for the class W to the case p 6= 2, in the same way as the results obtained in Chapter for the class E. Of course, one way to do that is to prove the estimate Z 1/2 −1/2 |Kz(ζ)| ω(ζ) dA(ζ) . ω(z) . (5.1) D This condition will follow if one is able to show the following pointwise estimate for reproducing kernels: for each M ≥ 1, there exists a constant C > 0 (depending on M) such that, for each z, ξ ∈ D one has

1 1 min(τ(z), τ(ξ))M |K (ξ)| ≤ C ω(z)−1/2ω(ξ)−1/2 . (5.2) z τ(z) τ(ξ) |z − ξ|

The obtention of this estimate leads to the estimate (5.1), and according to the results of Chapter 4 to estimate the norm of the reproducing kernels in Ap(ωp/2) for p ≥ 1. It also leads to the same estimate even for 0 < p < 1 and even to estimate the p-norms p p/2 of Kz on the associated Bergman spaces A (ω∗ ) for associated weights ω∗ of the form α ω∗(z) = ω(z) τ(z) with α ∈ R. This will allow to study the case 0 < p < 1 and the associated weighted Bergman spaces.

97 In view of the estimates for the test functions Fa,p given in Lemma C, it seems reasonable to think that the pointwise estimate (5.2) must be true for weights in the class W, but we have not succeeded on that task (we have not been able to ﬁnd a reasonable condition for τ(z) in order to apply the methods developed by Marzo and Ortega-Cerd`ain [44] for weighted Fock spaces).

5.2 Localization and Compactness

In the recent years, a great effort has been done on looking for general conditions for describing the compactness of an operator acting either on standard Bergman spaces or in the classical Fock spaces [30, 31, 46, 66, 70]. In a recent paper, J. Xia and D. Zheng [70] introduced a class of “sufficiently localized” operators on the Fock space including the algebraic closure of the Toeplitz operators. They proved that every bounded operator T from the C∗-algebra generated by the sufficiently localized operators whose Berezin transform vanishes at infinity is compact in the Fock space. The concept of sufficiently localized operators was extended to a larger class of operators (either in the Fock space setting or in the Bergman space setting) by J. Isralowitz, M. Mitkowski and B. Wick [31] introducing what they called “weakly localized operators”. The analogue of this last concept is what we are going to introduce next on the setting of large weighted Bergman spaces, when the weight ω is in the class E considered in Chapter 4. For the class W, we need to restrict to the case p = 2 because of the lack of estimates for kKzkAp(ωp/2).

0 For 1 < p < ∞, let p denote its conjugate exponent. Let kp,z denote the normalized p p/2 reproducing kernels on A (ω ), that is, kp,z = Kz/kKzkAp(ωp/2). Given an operator T acting on Ap(ωp/2), we deﬁne the quantities Z dA(ξ) 0 W1(T ) := sup |hT kp,z, kp ,ξiω| 2 , z∈D D τ(ξ) and Z dA(z) 0 W2(T ) := sup |hT kp,z, kp ,ξiω| 2 . ξ∈D D τ(z) Deﬁnition 5.1. Let 1 < p < ∞. A linear operator T acting on Ap(ωp/2) is said to be weakly localized if W (T ) = max W1(T ),W2(T ) < ∞ and Z m dA(ξ) 0 V1 (T ) := sup |hT kp,z, kp ,ξiω| 2 → 0, c z∈D Dm(z) τ(ξ) Z m dA(z) 0 V2 (T ) := sup |hT kp,z, kp ,ξiω| 2 → 0, c ξ∈D Dm(ξ) τ(z) |z−ξ| m as m → ∞, where Dm(z) := ξ ∈ D : dτ (z, ξ) = min(τ(z),τ(ξ)) < 2 δ .

98 It is not diﬃcult to see that if W (T ) < ∞, then T is bounded on Ap(ωp/2). Thus the class of weakly localized operators is included in the class all bounded operators.

Lemma 5.1. Let ω ∈ E and 1 < p < ∞. If W (T ) < ∞, then T is bounded on Ap(ωp/2).

Proof. Let E be the linear span of the reproducing kernels of A2(ω), that is dense in Ap(ωp/2). If f ∈ E, by the reproducing formula, we have Z ∗ ∗ T f(z) = hT f, Kziω = hf, T Kziω = f(ξ) T Kz(ξ) ω(ξ) dA(ξ). D Here T ∗ denotes the adjoint operator, acting now on Ap0 (ωp0/2) (see the duality result in Theorem 4.8). Since

∗ T Kz(ξ) = kKξkAp(ωp/2) · kKzkAp0 (ωp0/2)hT kp,ξ, kp0,ziω and, because of Lemma 4.2,

Z p p p/2 kT fkAp(ωp/2) = |T f(z)| ω(z) dA(z) D

Z Z p−1 p p/2 dA(z) 0 . W2(T ) |f(ξ)| ω(ξ) |hT kp,ξ, kp ,ziω| 2 dA(ξ) D D τ(z)

p−1 p . W1(T ) · W2(T ) kfkAp(ωp/2).

This ﬁnishes the proof.

p p/2 Denote by Aω,p(D) the set of all weakly localized operators acting on A (ω ).

99 The main goal of this project is to extend to our large Bergman spaces, the results proved in [31] for standard Bergman spaces or classical Fock spaces. That is, we want to p p/2 show that if T ∈ Aω,p(D), then T is compact on A (ω ) if and only if BωT (z) → 0 as − |z| → 1 , where Bω denotes the ω-Berezin transform of the operator T given by

TKz(z) BωT (z) = 2 , z ∈ D. kKzk2 We have just begin to look at that problem, and following the lines of [31] we have proved ∞ that for ω ∈ E and 1 ≤ p < ∞, the class Aω,p(D) forms an algebra, and that for ϕ ∈ L (D), the Toeplitz operator Tϕ is in Aω,p(D). However, there is still plenty of work to do before getting the result we are looking for.

5.3 Big Hankel operators

Another interesting problem is to study, for weights ω in the class E, the simultaneous p p/2 boundedness and compactness of the big Hankel operators Hf and Hf acting on A (ω ) 1 1/2 for general symbols f satisfying fKz ∈ L (ω ) for each z ∈ D. We have been able to prove p p p/2 that, for 1 ≤ p < ∞, if f ∈ BMO (τ) then Hf and Hf are both bounded on A (ω ). p p For 1 ≤ p < ∞, and δ ∈ (0, mτ ), let BMOδ (τ) denote the space of Lloc-integrable functions g on D such that Z 1/p 1 p kgk p := sup |g(s) − gδ(z)| dA(s) , BMOδ (τ) 2 2 b z∈D δ τ(z) D(δτ(z)) where 1 Z g (z) := g(s) dA(s). bδ 2 2 δ τ(z) D(δτ(z)) We have been able to study the structure of these spaces, and in particular, they are p p independent of δ, so that we can ﬁx some δ ∈ (0, mτ ) and put BMO (τ) = BMOδ (τ). p Conversely, we conjecture that if Hf and Hf are both bounded, then f ∈ BMO (τ). There are several things supporting this conjecture: the analogue problem in the case of standard weighted Bergman spaces is true [77]; in case that the symbol g is analytic, then it is not diﬃcult to see that g being in BMOp(τ) is equivalent to τg0 being a bounded function and this is just the condition characterizing the boundedness of Hg obtained in Theorem 5.3. Finally, as we commented before, we know how to prove the result in case that the reproducing kernels Kz does not have any zero in D, but it is unlikely that this in going to happen in general.

100 5.4 Small Hankel operators

Another interesting operator is the small Hankel operator hg for a given holomorphic symbol g, deﬁned as Z hgf(z) = (gf)(ζ) Kz(ζ) ω(ζ) dA(ζ). D 2 2 A good problem for the future is to look at the boundedness of hg : A (ω) → A (ω). This problem is equivalent to characterizing the boundedness of the bilinear form Bg(f, h) = 2 2 hfh, giω acting on A (ω) × A (ω). It looks that new ideas must be developed in order to attach this problem. In case that one succeeds on ﬁnding such a characterization, then one can try to extend the results to Ap(ωp/2).

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